Using Logarithm Properties - Algebra 2

Which expression is equivalent to $\log_2\left(\frac{\sqrt[3]{x}}{y^2\sqrt{z}}\right)$ ?
Answer: $\frac{1}{3}\log_2(x) - 2\log_2(y) - \frac{1}{2}\log_2(z)$ When you encounter a logarithm of a complex expression with fractions and roots, your goal is to break it down using logarithm properties: the quotient rule, product rule, and power rule. Start by rewriting the expression using exponents instead of radical notation: $\log_2\left(\frac{\sqrt[3]{x}}{y^2\sqrt{z}}\right) = \log_2\left(\frac{x^{1/3}}{y^2 \cdot z^{1/2}}\right)$ Now apply the quotient rule: $\log_b\left(\frac{A}{B}\right) = \log_b(A) - \log_b(B)$ This gives you: $\log_2(x^{1/3}) - \log_2(y^2 \cdot z^{1/2})$ For the second term, use the product rule: $\log_b(A \cdot B) = \log_b(A) + \log_b(B)$ So: $\log_2(x^{1/3}) - [\log_2(y^2) + \log_2(z^{1/2})]$ Finally, apply the power rule: $\log_b(A^n) = n\log_b(A)$ This yields: $\frac{1}{3}\log_2(x) - 2\log_2(y) - \frac{1}{2}\log_2(z)$ Looking at the wrong answers: Choice A has a positive $\frac{1}{2}\log_2(z)$ term, which ignores that $\sqrt{z}$ is in the denominator and should be subtracted. Choice C incorrectly applies the power rule, using 3 instead of $\frac{1}{3}$ for the cube root. Choice D makes the same sign error as A and also incorrectly adds the $y^2$ term instead of subtracting it. Study tip: Always convert roots to fractional exponents first, then systematically apply quotient, product, and power rules in that order. Pay careful attention to signs when terms are in denominators.
Use logarithm properties to evaluate without a calculator: $\log_3\left(\frac{81}{9}\right)$
Answer: $2$ This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions into sums and differences or condense multiple logs into a single logarithm. The three logarithm properties come from exponent rules and work because logarithms ARE exponents: (1) Product property: $\log_b$ (xy) = $\log_b$ (x) + $\log_b$ (y) because multiplying numbers means adding their exponents ( $b^m$ × $b^n$ = $b^{m+n}$ ), (2) Quotient property: $\log_b$ (x/y) = $\log_b$ (x) - $\log_b$ (y) because dividing means subtracting exponents ( $b^m$ ÷ $b^n$ = $b^{m-n}$ ), (3) Power property: $\log_b$ ( $x^p$ ) = p· $\log_b$ (x) because raising to power means multiplying exponents ( $(b^m)^p$ = $b^{mp}$ ). These properties transform multiplication into addition, division into subtraction, and exponentiation into multiplication—powerful simplification tools! To evaluate log₃(81/9) without a calculator: (1) First simplify the fraction: 81/9 = 9. (2) So log₃(81/9) = log₃(9). (3) Recognize that 9 = 3², so log₃(9) = log₃(3²) = 2. Alternatively, use the quotient property: log₃(81/9) = log₃(81) - log₃(9) = log₃(3⁴) - log₃(3²) = 4 - 2 = 2. Both methods give the same answer! Choice A correctly evaluates the logarithm as 2. Choice C incorrectly just uses the exponent from 81 = 3⁴ without considering the division by 9. The quotient 81/9 simplifies to 9 = 3², not 81 = 3⁴. You must perform the division inside the logarithm or use the quotient property correctly! Evaluation strategy using properties: (1) Simplify the argument first if possible: 81/9 = 9. (2) Express result as power of base: 9 = 3². (3) Use $\log_b$ ( $b^n$ ) = n directly: log₃(3²) = 2. (4) Or use quotient property: log₃(81) - log₃(9) = 4 - 2 = 2. This flexibility in approach helps verify your answer—if two different methods give the same result, you can be confident it's correct!
Use log properties to expand the expression $\log_7\left(\frac{\sqrt{x}\,y^5}{w^3}\right)$ as a sum/difference of logarithms with coefficients. (Use $\sqrt{x}=x^{1/2}$ ).
Answer: $\tfrac{1}{2}\log_7(x)+5\log_7(y)-3\log_7(w)$ This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions into sums and differences or condense multiple logs into a single logarithm. The three logarithm properties come from exponent rules: (1) Product property: $log_b(xy) = log_b(x) + log_b(y)$ , (2) Quotient property: $log_b(x/y) = log_b(x) - log_b(y)$ , (3) Power property: $log_b(x^p) = p \log_b(x)$ , and remember that $\sqrt{x} = x^{1/2}$ . To expand $log_7(\sqrt{x} \cdot y^5 / w^3)$ : (1) Apply quotient property first: $log_7(\sqrt{x} \cdot y^5) - log_7(w^3)$ , (2) Apply product property to numerator: $log_7(\sqrt{x}) + log_7(y^5) - log_7(w^3)$ , (3) Apply power property to each term: $(1/2)log_7(x) + 5log_7(y) - 3log_7(w)$ , since $\sqrt{x} = x^{1/2}$ . Choice A correctly expands with fractional coefficient 1/2 for the square root term and integer coefficients for the other powers. Choice C incorrectly leaves the powers inside the logarithms—the power property requires bringing ALL exponents out front as coefficients, including fractional exponents from roots! Handling roots in logarithms: (1) Convert roots to fractional exponents: $\sqrt{x} = x^{1/2}$ , $\sqrt[3]{x} = x^{1/3}$ , etc., (2) Apply power property: $log_b(x^{1/2}) = (1/2)log_b(x)$ , (3) Don't be afraid of fractional coefficients—they're perfectly valid! The key insight: roots are just fractional powers, so they follow the same power property rule, giving fractional coefficients when expanded.
Condense to a single logarithm: $3\ln(u)+\ln(v)-2\ln(w)$ (Assume $u>0$ , $v>0$ , and $w>0$ .)
Answer: $\ln\left(\frac{u^3 v}{w^2}\right)$ This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions into sums and differences or condense multiple logs into a single logarithm. The three logarithm properties come from exponent rules and work because logarithms ARE exponents: (1) Product property: $\log_b$ (xy) = $\log_b$ (x) + $\log_b$ (y) because multiplying numbers means adding their exponents ( $b^m$ × $b^n$ = $b^{m+n}$ ), (2) Quotient property: $\log_b$ (x/y) = $\log_b$ (x) - $\log_b$ (y) because dividing means subtracting exponents ( $b^m$ ÷ $b^n$ = $b^{m-n}$ ), (3) Power property: $\log_b$ ( $x^p$ ) = p· $\log_b$ (x) because raising to power means multiplying exponents ( $(b^m)^p$ = $b^{mp}$ ). These properties transform multiplication into addition, division into subtraction, and exponentiation into multiplication—powerful simplification tools! To condense 3ln(u) + ln(v) - 2ln(w), we reverse the properties: (1) Apply power property backwards: 3ln(u) becomes ln(u³) and -2ln(w) becomes -ln(w²). (2) Now we have ln(u³) + ln(v) - ln(w²). (3) Apply product property backwards to addition: ln(u³) + ln(v) = ln(u³v). (4) Apply quotient property backwards to subtraction: ln(u³v) - ln(w²) = ln(u³v/w²). Final condensed form: ln(u³v/w²)! Choice A correctly applies the logarithm properties in reverse order to condense into a single logarithm. Choice D incorrectly interprets the operations: there is NO logarithm property that allows you to write 3ln(u) + ln(v) as ln(u³ + v). The coefficient 3 becomes an exponent giving u³, and then u³ and v MULTIPLY (not add) inside the logarithm. Remember: log(a) + log(b) = log(ab), NOT log(a+b)! Condensing strategy (reverse process): (1) Use power property backwards: coefficient in front becomes exponent (3log(x) becomes log(x³)). (2) Identify additions: log(a) + log(b) condenses to log(ab) using product property backwards. (3) Identify subtractions: log(a) - log(b) condenses to log(a/b) using quotient property backwards. (4) Combine all: the result should be single log of one big expression. Work systematically—handle coefficients first, then combine logs two at a time!
Expand the expression using logarithm properties: $\log_{10}\left(\frac{100x^4}{y^2}\right)$ (Assume $x>0$ and $y>0$ .)
Answer: $2+4\log_{10}(x)-2\log_{10}(y)$ This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions into sums and differences or condense multiple logs into a single logarithm. The three logarithm properties come from exponent rules and work because logarithms ARE exponents: (1) Product property: $\log_b$ (xy) = $\log_b$ (x) + $\log_b$ (y) because multiplying numbers means adding their exponents ( $b^m$ × $b^n$ = $b^{m+n}$ ), (2) Quotient property: $\log_b$ (x/y) = $\log_b$ (x) - $\log_b$ (y) because dividing means subtracting exponents ( $b^m$ ÷ $b^n$ = $b^{m-n}$ ), (3) Power property: $\log_b$ ( $x^p$ ) = p· $\log_b$ (x) because raising to power means multiplying exponents ( $(b^m)^p$ = $b^{mp}$ ). These properties transform multiplication into addition, division into subtraction, and exponentiation into multiplication—powerful simplification tools! To expand log₁₀(100x⁴/y²) using properties: (1) Apply quotient property first: log₁₀(100x⁴) - log₁₀(y²) because the overall structure is a fraction. (2) Apply product property to numerator: log₁₀(100) + log₁₀(x⁴) - log₁₀(y²). (3) Apply power property to the terms with exponents: log₁₀(100) + 4log₁₀(x) - 2log₁₀(y). (4) Evaluate log₁₀(100) = log₁₀(10²) = 2. Final expanded form: 2 + 4log₁₀(x) - 2log₁₀(y). The properties break complex log into simple pieces! Choice B correctly applies the logarithm properties and evaluates log₁₀(100) = 2. Choice A incorrectly leaves log₁₀(100) unexpanded instead of evaluating it as 2, and leaves the powers inside as x⁴ and y² instead of bringing the exponents down as coefficients. When the base is 10 and you have log₁₀(100), you should recognize 100 = 10² and evaluate it! Expanding logarithm strategy: (1) Identify overall structure: is it product, quotient, or power? Start with outermost operation. (2) Apply matching property: multiplication becomes +, division becomes -, exponent becomes coefficient. (3) Work from outside in: handle fraction first (quotient property), then products within (product property), then exponents (power property). (4) Evaluate any logarithms you can: log₁₀(10) = 1, log₁₀(100) = 2, log₁₀(1000) = 3, etc. Recognizing powers of the base simplifies your final answer!
Use the logarithm properties $\log_b(xy)=\log_b(x)+\log_b(y),\quad \log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y),\quad \log_b(x^p)=p\log_b(x)$ to simplify the expression: $\frac{\log_5(125x)}{\log_5(5)}.$
Answer: $3+\log_5(x)$ This question tests your understanding of logarithm properties and the special case when the denominator is $\log_b$ (b) = 1, which allows for elegant simplification. The key insight is that $\log_5$ (5) = 1 because 5¹ = 5, so the expression becomes $\log_5$ (125x)/1 = $\log_5$ (125x). To simplify $\log_5$ (125x)/ $\log_5$ (5): First recognize that $\log_5$ (5) = 1, so we have $\log_5$ (125x)/1 = $\log_5$ (125x). Next, apply the product property: $\log_5$ (125x) = $\log_5$ (125) + $\log_5$ (x). Then evaluate $\log_5$ (125): Since 125 = 5³, we have $\log_5$ (125) = $\log_5$ (5³) = 3. Therefore: $\log_5$ (125x) = 3 + $\log_5$ (x). Choice B correctly shows the simplified form as 3 + $\log_5$ (x), recognizing that 125 = 5³ contributes a value of 3. Choice A just writes $\log_5$ (125x) without simplifying—while not wrong, it misses the opportunity to evaluate $\log_5$ (125) = 3. Choice C incorrectly suggests $\log_5$ (125) = $\log_5$ (3), but 125 = 5³, not 3·5. Simplification strategy when $\log_b$ (b) appears: (1) Always remember $\log_b$ (b) = 1 for any base b, (2) This often creates opportunities to simplify further, (3) Look for perfect powers of the base to evaluate (like 125 = 5³ here), (4) Express the final answer in simplest form by evaluating any logs of perfect powers. This type of problem rewards recognizing special values and powers of the base!
Condense to a single logarithm using $\log_b(xy)=\log_b(x)+\log_b(y),\quad \log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y),\quad \log_b(x^p)=p\log_b(x):$ $3\ln(x)-\ln(y)+2\ln(z).$
Answer: $\ln\left(\frac{x^3z^2}{y}\right)$ This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions into sums and differences or condense multiple logs into a single logarithm. The three logarithm properties work in reverse for condensing: (1) Power property backwards: $p \cdot \log_b(x)$ becomes $\log_b(x^p)$ , (2) Product property backwards: $\log_b(x) + \log_b(y)$ becomes $\log_b(xy)$ , (3) Quotient property backwards: $\log_b(x) - \log_b(y)$ becomes $\log_b(x/y)$ . To condense $3\ln(x) - \ln(y) + 2\ln(z)$ : (1) Apply power property backwards: $3\ln(x) = \ln(x^3)$ , $2\ln(z) = \ln(z^2)$ , giving $\ln(x^3) - \ln(y) + \ln(z^2)$ , (2) Rearrange to group positive terms: $\ln(x^3) + \ln(z^2) - \ln(y)$ , (3) Apply product property to positive terms: $\ln(x^3 \cdot z^2) - \ln(y)$ , (4) Apply quotient property: $\ln(x^3 z^2 / y)$ . Choice A correctly condenses into a single logarithm with $x^3 z^2$ in numerator and y in denominator. Choice B incorrectly puts y in the numerator—remember that $-\ln(y)$ means y goes in the denominator when condensing! Condensing strategy with mixed operations: (1) Convert all coefficients to exponents first using power property backwards, (2) Group positive logs (these will multiply) and negative logs (these will divide), (3) Apply product property to combine all positive terms into numerator, (4) Apply quotient property to put negative terms in denominator. Think of it as building a fraction: positive logs multiply in the numerator, negative logs multiply in the denominator!
Expand the logarithmic expression using the product, quotient, and power properties: $\log_3\left(\frac{x^2 y}{z^3}\right)$ (Assume $x>0$ , $y>0$ , and $z>0$ .)
Answer: $2\log_3(x)+\log_3(y)-3\log_3(z)$ This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions into sums and differences or condense multiple logs into a single logarithm. The three logarithm properties come from exponent rules and work because logarithms ARE exponents: (1) Product property: $\log_b$ (xy) = $\log_b$ (x) + $\log_b$ (y) because multiplying numbers means adding their exponents ( $b^m$ × $b^n$ = $b^{m+n}$ ), (2) Quotient property: $\log_b$ (x/y) = $\log_b$ (x) - $\log_b$ (y) because dividing means subtracting exponents ( $b^m$ ÷ $b^n$ = $b^{m-n}$ ), (3) Power property: $\log_b$ ( $x^p$ ) = p· $\log_b$ (x) because raising to power means multiplying exponents ( $(b^m)^p$ = $b^{mp}$ ). These properties transform multiplication into addition, division into subtraction, and exponentiation into multiplication—powerful simplification tools! To expand $\log_3$ (x²y/z³) using properties: (1) Apply quotient property first: $\log_3$ (x²y) - $\log_3$ (z³) because the overall structure is a fraction. (2) Apply product property to first log: $\log_3$ (x²) + $\log_3$ (y) because x²y is a product. (3) Apply power property to each: 2log_3(x) + $\log_3$ (y) - 3log_3(z) because x² has exponent 2 and z³ has exponent 3. Final expanded form: 2log_3(x) + $\log_3$ (y) - 3log_3(z). The properties break complex log into simple pieces! Choice B correctly applies the logarithm properties in the right order to expand into sum and difference of simpler logs. Choice A incorrectly leaves the exponents inside the logarithms instead of bringing them down as coefficients—the power property says log(x²) = 2log(x), not log(x²) stays as is. The exponents must come down in front as multipliers! Expanding logarithm strategy: (1) Identify overall structure: is it product, quotient, or power? Start with outermost operation. (2) Apply matching property: multiplication becomes +, division becomes -, exponent becomes coefficient. (3) Work from outside in: handle fraction first (quotient property), then products within (product property), then exponents (power property). (4) Continue until each log has single variable with coefficient. Example: log(x²y³/z) becomes log(x²y³) - log(z), then 2log(x) + 3log(y) - log(z). Systematic outside-in approach!
Condense to a single logarithm (base 10): $\log(x)-\tfrac{1}{2}\log(y)+\log(z)$
Answer: $\log\left(\frac{xz}{\sqrt{y}}\right)$ This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions into sums and differences or condense multiple logs into a single logarithm. The three properties in reverse (for condensing) are: (1) Power property backwards: p· $\log_b$ (x) becomes $\log_b$ ( $x^p$ ), (2) Product property backwards: $\log_b$ (x) + $\log_b$ (y) becomes $\log_b$ (xy), (3) Quotient property backwards: $\log_b$ (x) - $\log_b$ (y) becomes $\log_b$ (x/y)—these transform sums into products, differences into quotients, and coefficients into exponents! To condense log(x) - (1/2)log(y) + log(z): (1) Apply power property to middle term: -(1/2)log(y) = log( $y^{-1/2}$ ) = log(1/√y), (2) Rewrite expression: log(x) + log(1/√y) + log(z), but wait—we need to be careful with the negative! Actually, -(1/2)log(y) means we're subtracting, so we have log(x) - (1/2)log(y) + log(z), (3) Apply power property: (1/2)log(y) = log(√y), (4) Rearrange: log(x) + log(z) - log(√y), (5) Apply product then quotient: log(xz) - log(√y) = log(xz/√y). Choice A correctly shows log(xz/√y) as the condensed form. Choice D incorrectly has y² in the denominator instead of √y—the coefficient 1/2 becomes an exponent 1/2 (which is square root), not 2! Remember that fractional exponents represent roots: $x^{1/2}$ = √x. Condensing strategy with fractional coefficients: (1) Convert fractional coefficients to fractional exponents carefully: (1/2)log(x) = log( $x^{1/2}$ ) = log(√x), (2) Pay attention to signs: negative means subtraction will lead to division, (3) Group additions first (they become products), then handle subtractions (they become quotients), (4) Fractional exponents mean roots: 1/2 → square root, 1/3 → cube root. This systematic approach prevents common errors with roots and fractions!
Express using the change-of-base formula (in terms of natural logs): $\log_5(20)$
Answer: $\dfrac{\ln(20)}{\ln(5)}$ This question tests your understanding of the change-of-base formula, which allows you to express logarithms with any base in terms of logarithms with a more convenient base (like base 10 or base e). The change-of-base formula states: $\log_b$ (x) = $\log_c$ (x)/ $\log_c$ (b) for any valid base c. This works because if $b^y$ = x, then taking $\log_c$ of both sides gives y· $\log_c$ (b) = $\log_c$ (x), so y = $\log_c$ (x)/ $\log_c$ (b). Most commonly, we use natural logarithm (ln) or common logarithm (log₁₀) as the new base because calculators have these functions built in! To express log₅(20) using natural logs: Apply the change-of-base formula with c = e: log₅(20) = ln(20)/ln(5). This converts the base-5 logarithm into a ratio of natural logarithms that can be evaluated on any scientific calculator. The formula essentially asks: "If $5^y$ = 20, what is y?" and solves it using natural logs. Choice B correctly applies the change-of-base formula as ln(20)/ln(5). Choice A incorrectly inverts the fraction—it shows ln(5)/ln(20), but the correct formula has the argument (20) in the numerator and the base (5) in the denominator. Remember: log_base(argument) = ln(argument)/ln(base). The argument goes on top! Change-of-base strategy: (1) Identify what you're converting: $\log_b$ (x) where b is the base and x is the argument. (2) Choose your new base (usually e for ln or 10 for log). (3) Apply formula: $\log_b$ (x) = ln(x)/ln(b) or $\log_b$ (x) = log(x)/log(b). (4) The argument x goes in numerator, base b goes in denominator—don't mix them up! This formula is essential for calculating logarithms with unusual bases on a calculator that only has ln and log₁₀ buttons.
Use the logarithm properties $\log_b(xy)=\log_b(x)+\log_b(y),\quad \log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y),\quad \log_b(x^p)=p\log_b(x)$ to write the expression as a single logarithm: $3\log_5(a)-\tfrac12\log_5(b)+\log_5(c).$
Answer: $\log_5\left(\dfrac{a^3c}{\sqrt{b}}\right)$ This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you condense multiple logarithms into a single logarithm expression. The three logarithm properties come from exponent rules: (1) Product property: $\log_b$ (x) + $\log_b$ (y) = $\log_b$ (xy) because adding exponents means multiplying the original numbers, (2) Quotient property: $\log_b$ (x) - $\log_b$ (y) = $\log_b$ (x/y) because subtracting exponents means dividing, (3) Power property: p· $\log_b$ (x) = $\log_b$ ( $x^p$ ) because a coefficient in front becomes an exponent inside. To condense 3log_5(a) - ½ $\log_5$ (b) + $\log_5$ (c): First apply power property backwards to move coefficients up as exponents: $\log_5$ (a³) - $\log_5$ ( $b^{1/2}$ ) + $\log_5$ (c). Note that $b^{1/2}$ = √b. Next, work left to right combining terms: $\log_5$ (a³) - $\log_5$ (√b) becomes $\log_5$ (a³/√b) using quotient property. Finally add the last term: $\log_5$ (a³/√b) + $\log_5$ (c) becomes $\log_5$ ((a³/√b)·c) = $\log_5$ (a³c/√b) using product property. Choice A correctly condenses the expression into a single logarithm with a³c in the numerator and √b in the denominator. Choice D incorrectly tries to combine logs using addition/subtraction of the arguments (a³ - √b + c), but there is NO logarithm property for log(x + y) or log(x - y)—you can only combine logs when they're being added (→ multiplication) or subtracted (→ division). Condensing strategy: (1) First convert all coefficients to exponents using power property backwards, (2) Then combine logs two at a time using product/quotient properties, (3) Addition of logs → multiplication inside, subtraction of logs → division inside. Remember: coefficients become exponents, and you can only combine logs through multiplication/division, never through addition/subtraction of arguments!
Use the logarithm properties $\log_b(xy)=\log_b(x)+\log_b(y),\quad \log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y),\quad \log_b(x^p)=p\log_b(x)$ to expand the expression $\log_7\left(\frac{\sqrt{x}\,y^3}{w^2}\right).$
Answer: $\tfrac12\log_7(x)+3\log_7(y)-2\log_7(w)$ This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions containing roots and powers into sums and differences of simpler logs. The three properties are: (1) Product property: $\log_b$ (xy) = $\log_b$ (x) + $\log_b$ (y), (2) Quotient property: $\log_b$ (x/y) = $\log_b$ (x) - $\log_b$ (y), (3) Power property: $\log_b$ ( $x^p$ ) = p· $\log_b$ (x), and remember that √x = $x^{1/2}$ . To expand $\log_7$ (√x·y³/w²): First apply quotient property to the fraction: $\log_7$ (√x·y³) - $\log_7$ (w²). Next apply product property to the numerator: $\log_7$ (√x) + $\log_7$ (y³) - $\log_7$ (w²). Then apply power property to each term, remembering √x = $x^{1/2}$ : (1/2) $\log_7$ (x) + 3log_7(y) - 2log_7(w). Choice A correctly shows the fully expanded form with the square root converted to the coefficient 1/2, the cube to coefficient 3, and the square to coefficient 2. Choice D incorrectly tries to write √x + y³ inside a single log, but this violates the product property—log(√x·y³) expands to log(√x) + log(y³), not log(√x + y³). There's no logarithm property for addition inside! The key insight for roots: √x = $x^{1/2}$ , so log(√x) = log( $x^{1/2}$ ) = (1/2)log(x). Expansion strategy with roots and powers: (1) Recognize roots as fractional exponents (√x = $x^{1/2}$ , ∛x = $x^{1/3}$ ), (2) Apply properties systematically from outside in, (3) All exponents become coefficients, including fractions from roots. This creates a clean expanded form where each log contains a single variable and all powers appear as coefficients out front!
If $\log_6(m) = 2$ and $\log_6(n) = -1$ , what is $\log_6\left(\frac{m^2\sqrt{n}}{36}\right)$ ?
Answer: $1.5$ Using logarithm properties: $\log_6\left(\frac{m^2\sqrt{n}}{36}\right) = \log_6(m^2) + \log_6(\sqrt{n}) - \log_6(36) = 2\log_6(m) + \frac{1}{2}\log_6(n) - \log_6(6^2) = 2(2) + \frac{1}{2}(-1) - 2 = 4 - 0.5 - 2 = 1.5$ . Choice A results from incorrectly calculating $4 - 0.5 - 1 = 2.5$ (using $\log_6(36) = 1$ ). Choice B results from omitting the $\log_6(36)$ term entirely. Choice D results from incorrectly calculating $\log_6(36) = 1$ instead of 2.
Simplify using logarithm properties: $\log_4(64) - \log_4(4)$
Answer: $2$ This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—to simplify differences of logs into a single value. The three logarithm properties come from exponent rules and work because logarithms are exponents: (1) Product property: $\log_b(xy) = \log_b(x) + \log_b(y)$ , (2) Quotient property: $\log_b(x) - \log_b(y) = \log_b(x/y)$ , (3) Power property: $\log_b(x^p) = p \log_b(x)$ . To simplify $\log_4(64) - \log_4(4)$ , apply quotient: $\log_4(64/4) = \log_4(16)$ ; since $16 = 4^2$ , $\log_4(16) = 2$ . Choice B correctly uses the quotient property and evaluates by rewriting as a power of the base. Choice A might come from subtracting the arguments directly—remember to combine logs first! Simplify by combining with properties before evaluating, and express arguments as powers of the base when possible. This makes numerical answers pop out easily—excellent effort!
Use the properties $\log_b(xy)=\log_b(x)+\log_b(y),\quad \log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y),\quad \log_b(x^p)=p\log_b(x)$ to expand $\log\left(\frac{(mn^3)^2}{p}\right)$ , where $\log$ is base 10. Write your answer as a sum/difference of logs with coefficients.
Answer: $2\log(m)+6\log(n)-\log(p)$ This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions into sums and differences or condense multiple logs into a single logarithm. The three logarithm properties come from exponent rules: (1) Product property: $\log_b$ (xy) = $\log_b$ (x) + $\log_b$ (y), (2) Quotient property: $\log_b$ (x/y) = $\log_b$ (x) - $\log_b$ (y), (3) Power property: $\log_b$ ( $x^p$ ) = p· $\log_b$ (x). To expand log( $(mn^3)^2$ /p): (1) Apply quotient property first: log( $(mn^3)^2$ ) - log(p), (2) Apply power property to $(mn^3)^2$ : log( $(mn^3)^2$ ) = 2log( $mn^3$ ), so we have 2log( $mn^3$ ) - log(p), (3) Apply product property inside: 2[log(m) + log( $n^3$ )] - log(p) = 2log(m) + 2log( $n^3$ ) - log(p), (4) Apply power property to $n^3$ : 2log(m) + 2·3log(n) - log(p) = 2log(m) + 6log(n) - log(p). Choice A correctly applies all properties in the right order, carefully handling the outer exponent 2 that applies to the entire product $mn^3$ . Choice D incorrectly suggests log(m + $n^3$ ), but there is NO property for log(x + y)—the product property applies to multiplication log(xy), not addition! Expanding with nested exponents: (1) Handle outer operations first (quotient in this case), (2) When you see (expression)^p, use power property to get p·log(expression), (3) Then expand what's inside the log using product property, (4) Distribute any coefficients from outer power property to all terms inside. The key insight: $(mn^3)^2$ = m^2n^6, so the final coefficients are 2 for m and 6 for n!
Write as a single logarithm (base 5): $\log_5(m)-2\log_5(n)+\log_5(p).$
Answer: $\log_5\left(\frac{mp}{n^2}\right)$ This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you condense multiple logs into a single logarithm. The three logarithm properties come from exponent rules and work because logarithms are exponents: (1) Product property: $\log_b$ (x) + $\log_b$ (y) = $\log_b$ (xy), (2) Quotient property: $\log_b$ (x) - $\log_b$ (y) = $\log_b$ (x/y), (3) Power property: p $\log_b$ (x) = $\log_b$ ( $x^p$ ). To condense $\log_5$ (m) - 2 $\log_5$ (n) + $\log_5$ (p), first power on the middle: $\log_5$ (m) - $\log_5$ ( $n^2$ ) + $\log_5$ (p), then combine additions and subtraction: $\log_5$ ( (m p) / $n^2$ ). Choice A correctly condenses using power, product, and quotient in reverse. Choice B flips the signs incorrectly—subtraction means denominator, not numerator! For condensing, apply power first to coefficients, then products for + and quotients for -. This reverse expansion technique is powerful—nice work!
If $\log_5(a) + 3\log_5(b) - \log_5(c) = \log_5(20)$ , which equation is equivalent using logarithm properties?
Answer: $\frac{ab^3}{c} = 20$ When you encounter logarithmic equations with multiple terms, your goal is to use logarithm properties to simplify the left side and then apply the definition of logarithms to eliminate the log notation entirely. Start by applying the key logarithm properties to the left side of $\log_5(a) + 3\log_5(b) - \log_5(c) = \log_5(20)$ . First, use the power property: $3\log_5(b) = \log_5(b^3)$ . Next, apply the addition property (logs of products) and subtraction property (logs of quotients): $\log_5(a) + \log_5(b^3) - \log_5(c) = \log_5\left(\frac{a \cdot b^3}{c}\right)$ . Now your equation becomes $\log_5\left(\frac{ab^3}{c}\right) = \log_5(20)$ . Since both sides have the same base logarithm, the arguments must be equal: $\frac{ab^3}{c} = 20$ . Looking at the wrong answers: Choice A ( $ab^3 - c = 20$ ) incorrectly treats subtraction of logs as regular subtraction instead of division. Choice B ( $a + 3b - c = 20$ ) makes the fundamental error of treating logarithms as if they follow normal arithmetic rules, ignoring all logarithm properties. Choice C ( $\frac{ab^3}{c} = 5^{20}$ ) correctly applies logarithm properties but fails to recognize that when $\log_5(x) = \log_5(20)$ , then $x = 20$ , not $x = 5^{20}$ . Remember: when working with logarithmic equations, always consolidate terms using properties first, then use the fact that if $\log_b(x) = \log_b(y)$ , then $x = y$ .
Use the logarithm properties $\log_b(xy)=\log_b(x)+\log_b(y),\quad \log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y),\quad \log_b(x^p)=p\log_b(x)$ to write as a single logarithm: $2\ln(x)+\ln(y)-3\ln(z).$
Answer: $\ln\left(\dfrac{x^2y}{z^3}\right)$ This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you condense multiple logarithms into a single logarithm expression. The properties work backwards for condensing: (1) Power property: p·ln(x) = ln( $x^p$ ) moves coefficients up as exponents, (2) Product property: ln(x) + ln(y) = ln(xy) turns addition into multiplication, (3) Quotient property: ln(x) - ln(y) = ln(x/y) turns subtraction into division. To condense 2ln(x) + ln(y) - 3ln(z): First apply the power property to move coefficients up: ln(x²) + ln(y) - ln(z³). Next combine the first two terms using product property: ln(x²y) - ln(z³). Finally apply quotient property for the subtraction: ln(x²y/z³). Choice A correctly shows the condensed form as a single natural logarithm of the fraction x²y/z³. Choice B incorrectly interprets the coefficient 3 in front of ln(z) as going inside the log as 3z instead of as an exponent z³—the power property says 3ln(z) = ln(z³), not ln(3z)! This is a critical distinction: coefficients become exponents, not multipliers inside. Condensing strategy: (1) Convert all coefficients to exponents first using power property, (2) Identify additions (will become products) and subtractions (will become quotients), (3) Combine terms systematically, usually left to right, (4) The result should be a single log of one expression. Remember the pattern: coefficient → exponent, addition → multiplication, subtraction → division!
Use the product, quotient, and power properties of logarithms to write the following as a single logarithm (base 10): $2\log(x) + \frac{1}{2}\log(y) - 3\log(z).$
Answer: $\log\left(\frac{x^2 y^{1/2}}{z^3}\right)$ This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you condense multiple logarithmic expressions into a single logarithm. The three logarithm properties come from exponent rules and work because logarithms ARE exponents: (1) Product property: $\log_b(xy) = \log_b(x) + \log_b(y)$ because multiplying numbers means adding their exponents, (2) Quotient property: $\log_b(x/y) = \log_b(x) - \log_b(y)$ because dividing means subtracting exponents, (3) Power property: $\log_b(x^p) = p \log_b(x)$ because raising to power means multiplying exponents. These properties transform addition into multiplication, subtraction into division, and coefficients into exponents—powerful simplification tools! For condensing $2\log(x) + \frac{1}{2}\log(y) - 3\log(z)$ , reverse the properties: (1) Apply power property backwards: $\log(x^2) + \log(y^{1/2}) - \log(z^3)$ by bringing coefficients up as exponents. (2) Apply product property backwards to the sum: $\log(x^2 y^{1/2}) - \log(z^3)$ because sum of logs becomes log of product. (3) Apply quotient property backwards: $\log\left( \frac{x^2 y^{1/2}}{z^3} \right)$ because difference becomes quotient. Final condensed form: $\log\left( \frac{x^2 \sqrt{y}}{z^3} \right)$ . Choice A correctly applies the logarithm properties in the right order to condense into a single logarithm. Choice B incorrectly ignores the 1/2 exponent on y, treating it as $\log(y)$ instead of $ \log( $\sqrt{y}$ ) $—always remember to apply the power rule first to coefficients! Condensing strategy: (1) Use power property forward: coefficient in front becomes exponent (like $2\log(x)$ to $ \log( $x^2$ ) $). (2) Identify additions: $\log(a) + \log(b)$ condenses to $\log(a b)$ . (3) Identify subtractions: $\log(a) - \log(b)$ condenses to $\log(a/b)$ . (4) Combine all step-by-step into one log. Great job—keep building your skills with this reverse process!
Use logarithm properties to evaluate without a calculator: $\log_2(8\cdot 32)$
Answer: $8$ This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions into sums and differences or condense multiple logs into a single logarithm. The three logarithm properties come from exponent rules and work because logarithms ARE exponents: (1) Product property: $\log_b$ (xy) = $\log_b$ (x) + $\log_b$ (y) because multiplying numbers means adding their exponents ( $b^m$ × $b^n$ = $b^{m+n}$ ), (2) Quotient property: $\log_b$ (x/y) = $\log_b$ (x) - $\log_b$ (y) because dividing means subtracting exponents ( $b^m$ ÷ $b^n$ = $b^{m-n}$ ), (3) Power property: $\log_b$ ( $x^p$ ) = p· $\log_b$ (x) because raising to power means multiplying exponents ( $(b^m)^p$ = $b^{mp}$ ). These properties transform multiplication into addition, division into subtraction, and exponentiation into multiplication—powerful simplification tools! To evaluate log₂(8·32) without a calculator: (1) First recognize that 8 = 2³ and 32 = 2⁵, so 8·32 = 2³·2⁵ = 2⁸ = 256. (2) Therefore log₂(8·32) = log₂(256) = log₂(2⁸) = 8, because log₂ asks "2 to what power gives 256?" and the answer is 8. Alternatively, use the product property: log₂(8·32) = log₂(8) + log₂(32) = log₂(2³) + log₂(2⁵) = 3 + 5 = 8. Both methods give the same answer! Choice B correctly evaluates the logarithm as 8. Choice C incorrectly multiplies the individual logarithms: log₂(8) = 3 and log₂(32) = 5, but 3×5 = 15 is wrong. The product property says to ADD logarithms when the arguments are multiplied: log(xy) = log(x) + log(y), not multiply them! Evaluation strategy using properties: (1) Express numbers as powers of the base when possible: 8 = 2³, 32 = 2⁵. (2) Apply product property: $\log_b$ (xy) = $\log_b$ (x) + $\log_b$ (y). (3) Use the fact that $\log_b$ ( $b^n$ ) = n directly. (4) Add the exponents: log₂(2³) + log₂(2⁵) = 3 + 5 = 8. This method avoids calculating large products! Remember: when arguments multiply, logarithms ADD—this is the key insight that makes logarithms useful for simplifying calculations with large numbers.

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