Logarithm Properties - Math 3

If $\log_2(x) + \log_2(y) = 5$ and $\log_2(x) - \log_2(y) = 1$ , what is the value of $xy$ ?
Answer: 32 Using logarithm properties: From the given equations, $\log_2(xy) = \log_2(x) + \log_2(y) = 5$ , so $xy = 2^5 = 32$ . We can verify by solving the system: adding the equations gives $2\log_2(x) = 6$ , so $\log_2(x) = 3$ and $x = 8$ . Substituting back: $\log_2(y) = 2$ , so $y = 4$ . Thus $xy = 32$ . Choice A uses $2^4$ instead of $2^5$ . Choice C incorrectly calculates $x + y = 12$ then squares it. Choice D uses $2^7$ by incorrectly adding both original equations.
Given $\log_3(a) = 2$ and $\log_3(b) = -1$ , what is $\log_3\left(\frac{a^2}{b^3}\right)$ ?
Answer: $7$ When you encounter logarithmic expressions with given logarithm values, you'll need to apply the fundamental properties of logarithms to break down complex expressions into simpler parts. Start by using the logarithm properties: $\log_b(x^n) = n\log_b(x)$ and $\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)$ . This means $\log_3\left(\frac{a^2}{b^3}\right) = \log_3(a^2) - \log_3(b^3) = 2\log_3(a) - 3\log_3(b)$ . Now substitute the given values: $\log_3(a) = 2$ and $\log_3(b) = -1$ . This gives you $2\log_3(a) - 3\log_3(b) = 2(2) - 3(-1) = 4 - (-3) = 4 + 3 = 7$ . Looking at the wrong answers: Choice A ( $-1$ ) likely comes from incorrectly calculating $2(-1) - 3(2) = -5$ , mixing up the given values. Choice B ( $1$ ) might result from errors like $2(2) - 3(1) = 1$ , possibly confusing $\log_3(b) = -1$ with $\log_3(b) = 1$ . Choice D ( $5$ ) could come from sign errors, such as calculating $2(2) - 3(-1) = 4 - 3 = 1$ incorrectly, or $2(2) + 3(-1) = 4 - 3 = 1$ then adding instead of subtracting. The answer is C ( $7$ ). Study tip: Always write out the logarithm properties step-by-step before substituting values. This prevents sign errors and helps you track which operations to perform when dealing with quotients and exponents in logarithmic expressions.
Which statement best explains why $\log_a(xy) = \log_a(x) + \log_a(y)$ requires both $x > 0$ and $y > 0$ ?
Answer: Logarithms are only defined for positive real numbers in their standard form When working with logarithmic properties, you need to understand the fundamental domain restrictions that make these functions well-defined. The logarithm function $\log_a(x)$ exists only when its argument is a positive real number—this is a foundational requirement, not just a convenient restriction. The property $\log_a(xy) = \log_a(x) + \log_a(y)$ requires both $x > 0$ and $y > 0$ because logarithms themselves are only defined for positive real numbers in their standard form. If either $x$ or $y$ were negative or zero, then $\log_a(x)$ or $\log_a(y)$ would be undefined, making the entire equation meaningless. This makes answer C correct. Let's examine why the other options miss the mark. Answer A incorrectly focuses on the base $a$ , but the base isn't what's problematic here—it's the arguments $x$ and $y$ that must be positive. Answer B suggests the issue is purely algebraic about the product $xy$ , but even if $xy > 0$ (say, when both $x$ and $y$ are negative), you still couldn't evaluate $\log_a(x)$ or $\log_a(y)$ individually. Answer D incorrectly restricts the domain to positive integers, when logarithms work for all positive real numbers. Remember this key principle: before applying any logarithm property, check that every logarithm in the equation is actually defined. The domain of logarithms—positive real numbers only—often determines what values are permissible in logarithmic equations and properties.
Which equation demonstrates the correct application of the change of base formula to rewrite $\log_8(64)$ ?
Answer: $\log_8(64) = \frac{\log(64)}{\log(8)}$ When you encounter logarithms with bases other than 10 or e, the change of base formula becomes essential for solving problems with a calculator or converting between different forms. The change of base formula states that $\log_b(a) = \frac{\log(a)}{\log(b)}$ , where the logarithms on the right can be in any convenient base (typically base 10). To apply this to $\log_8(64)$ , you place 64 (the argument) in the numerator and 8 (the original base) in the denominator: $\log_8(64) = \frac{\log(64)}{\log(8)}$ . This is exactly what option C shows. Option A incorrectly applies the quotient rule for logarithms, which would give you $\log\left(\frac{64}{8}\right) = \log(8)$ , not $\log_8(64)$ . The quotient rule only works when both logarithms have the same base. Option B flips the fraction, putting the original base in the numerator and the argument in the denominator. This reversal would actually give you $\frac{1}{\log_{64}(8)}$ , which is completely different from what we want. Option D attempts to use the change of base formula but keeps the same base (8) for both the original expression and the denominator. This creates a circular definition that doesn't actually change the base at all. Remember the change of base formula as "new over old" - the original argument goes over the original base. This pattern will help you avoid the common trap of reversing the fraction, which appears frequently on math exams.
For which value of $k$ is the equation $\log_2(x + 1) + \log_2(x - 1) = \log_2(k)$ equivalent to $x^2 - 1 = k$ ?
Answer: All values of $k$ such that $k > 0$ and $x > 1$ Using the product rule: $\log_2(x + 1) + \log_2(x - 1) = \log_2((x + 1)(x - 1)) = \log_2(x^2 - 1)$ . So the equation becomes $\log_2(x^2 - 1) = \log_2(k)$ , which gives $x^2 - 1 = k$ . However, for the original equation to be valid, we need $x + 1 > 0$ and $x - 1 > 0$ , so $x > 1$ . This restricts the domain and thus the possible values of $k$ . Choice A ignores domain restrictions. Choice B considers $k > 0$ but misses the $x > 1$ constraint. Choice C incorrectly includes $k = 0$ .
If $\log_5(m) = p$ and $\log_5(n) = q$ , which expression represents $\log_5(\frac{25m^3}{n^2})$ ?
Answer: $2 + 3p - 2q$ Using logarithm properties: $\log_5(\frac{25m^3}{n^2}) = \log_5(25) + \log_5(m^3) - \log_5(n^2) = \log_5(5^2) + 3\log_5(m) - 2\log_5(n) = 2 + 3p - 2q$ . Choice B incorrectly treats $\log_5(25)$ as 25 instead of 2. Choice C misapplies the quotient rule as algebraic division. Choice D incorrectly applies the quotient rule to the variables $p$ and $q$ .
If $\log_2(x+3) - \log_2(x-1) = 2$ , then $x$ equals:
Answer: $\frac{7}{3}$ Using the quotient property: $\log_2\left(\frac{x+3}{x-1}\right) = 2$ . This means $\frac{x+3}{x-1} = 2^2 = 4$ . Cross-multiplying: $x + 3 = 4(x-1) = 4x - 4$ . Solving: $x + 3 = 4x - 4$ , so $7 = 3x$ , giving $x = \frac{7}{3}$ . We must check that $x > 1$ for the domain: $\frac{7}{3} \approx 2.33 > 1$ ✓. Choice A, B, and C result from arithmetic errors in solving the linear equation.
If $\log_6(18) + \log_6(2) = \log_6(k)$ , then $k$ equals:
Answer: $36$ Using the product property: $\log_6(18) + \log_6(2) = \log_6(18 \cdot 2) = \log_6(36)$ . Therefore $k = 36$ . Choice A comes from $18 + 2 = 20$ (incorrectly adding arguments), choice C from $18 \cdot 6 = 108$ (multiplying by base instead of the other argument), choice D from $6^3 = 216$ (misremembering that $6^2 = 36$ ).
If $\log_a(m) = p$ and $\log_a(n) = q$ , which expression represents $\log_a\left(\frac{m^3}{n^2\sqrt[4]{a}}\right)$ ?
Answer: $3p - 2q - \frac{1}{4}$ Using logarithm properties: $\log_a\left(\frac{m^3}{n^2\sqrt[4]{a}}\right) = \log_a(m^3) - \log_a(n^2) - \log_a(a^{1/4}) = 3\log_a(m) - 2\log_a(n) - \frac{1}{4}\log_a(a) = 3p - 2q - \frac{1}{4}$ . Choice B has wrong sign, choice C incorrectly uses division instead of subtraction, choice D uses $-4$ instead of $-\frac{1}{4}$ .
Given that $\log_3(x) = 2$ and $\log_9(y) = \frac{1}{2}$ , what is the value of $\log_3(\frac{xy}{3})$ ?
Answer: $2$ First, convert $\log_9(y) = \frac{1}{2}$ to base 3: Since $9 = 3^2$ , we have $\log_9(y) = \frac{\log_3(y)}{\log_3(9)} = \frac{\log_3(y)}{2} = \frac{1}{2}$ , so $\log_3(y) = 1$ . Now: $\log_3(\frac{xy}{3}) = \log_3(x) + \log_3(y) - \log_3(3) = 2 + 1 - 1 = 2$ . Choice A forgets to subtract $\log_3(3) = 1$ . Choice B adds all terms without subtracting. Choice D incorrectly computes $\log_3(y) = \frac{3}{2}$ .
If $\log_2(x) + \log_2(y) = 5$ and $\log_2(x) - \log_2(y) = 1$ , what is the value of $xy$ ?
Answer: 32 Using logarithm properties: $\log_2(x) + \log_2(y) = \log_2(xy) = 5$ , so $xy = 2^5 = 32$ . We can verify by solving the system: adding the equations gives $2\log_2(x) = 6$ , so $\log_2(x) = 3$ and $x = 8$ . Substituting back: $\log_2(8) + \log_2(y) = 5$ , so $3 + \log_2(y) = 5$ , giving $y = 4$ . Therefore $xy = 32$ .
If $\log_a(b) = 3$ and $\log_a(c) = -2$ , what is the value of $\log_a(\frac{b^2\sqrt{c}}{a})$ ?
Answer: $4$ Using logarithm properties: $\log_a(\frac{b^2\sqrt{c}}{a}) = \log_a(b^2) + \log_a(\sqrt{c}) - \log_a(a) = 2\log_a(b) + \frac{1}{2}\log_a(c) - 1 = 2(3) + \frac{1}{2}(-2) - 1 = 6 - 1 - 1 = 4$ . Choice A forgets to subtract $\log_a(a) = 1$ . Choice C incorrectly calculates $2(3) + \frac{1}{2}(-2) = 6 - 1 = 5$ then subtracts 2 instead of 1. Choice D makes an error in the power rule for $\sqrt{c}$ .
Which statement correctly justifies why $\log_2(a-b) = \log_2(a) - \log_2(b)$ is generally false?
Answer: The logarithm of a difference cannot be simplified using standard logarithm properties alone The logarithm of a difference $\log_2(a-b)$ cannot be simplified using the standard logarithm properties (product, quotient, power rules). These properties apply to products, quotients, and powers, not sums or differences of the arguments. Choice A confuses this with quotient property, choice C incorrectly limits logarithm properties to positive integers, choice D incorrectly suggests the false equation works under certain conditions.
Which expression is equivalent to $\log_3\left(\frac{9x^2}{\sqrt{y}}\right)$ when $x > 0$ and $y > 0$ ?
Answer: $2 + 2\log_3(x) - \frac{1}{2}\log_3(y)$ Using logarithm properties: $\log_3\left(\frac{9x^2}{\sqrt{y}}\right) = \log_3(9) + \log_3(x^2) - \log_3(\sqrt{y}) = \log_3(3^2) + 2\log_3(x) - \log_3(y^{1/2}) = 2 + 2\log_3(x) - \frac{1}{2}\log_3(y)$ . Choice B doesn't fully expand the power rule, choice C doesn't simplify $\log_3(9) = 2$ , and choice D has an incorrect sign.
Given that $\log_5(2) \approx 0.431$ , which value is closest to $\log_5(0.8)$ ?
Answer: $-0.097$ When you encounter logarithm problems with decimal inputs, the key is recognizing how to break down the number using logarithm properties, especially when you're given a related logarithm value. To find $\log_5(0.8)$ , start by rewriting 0.8 as a fraction: $0.8 = \frac{8}{10} = \frac{4}{5}$ . Now you can use the logarithm quotient rule: $\log_5(0.8) = \log_5\left(\frac{4}{5}\right) = \log_5(4) - \log_5(5)$ . Since $\log_5(5) = 1$ , you have $\log_5(0.8) = \log_5(4) - 1$ . Next, recognize that $4 = 2^2$ , so $\log_5(4) = \log_5(2^2) = 2\log_5(2)$ . Using the given value: $\log_5(4) = 2 \times 0.431 = 0.862$ . Therefore: $\log_5(0.8) = 0.862 - 1 = -0.138$ , which is closest to $-0.097$ . Option A (0.334) represents a common error of forgetting the negative sign and miscalculating the final arithmetic. Option B (0.097) makes the same arithmetic mistake as the correct answer but incorrectly assumes the result should be positive. Option C (-0.334) gets the negative sign right but likely confused the calculation, perhaps by incorrectly using $\log_5(2) - 1$ directly. Remember that logarithms of numbers between 0 and 1 are always negative, and practice breaking down decimal numbers into fractions to apply logarithm properties effectively.
If $\log_b(x) = 3\log_b(2) + \frac{1}{2}\log_b(9) - 2\log_b(3)$ , then $x$ equals: Give your answer in simplest form.
Answer: $\frac{8}{3}$ When you encounter logarithmic expressions with multiple terms, the key is using logarithm properties to simplify before solving. The three essential properties are: $n\log_b(a) = \log_b(a^n)$ , $\log_b(a) + \log_b(c) = \log_b(ac)$ , and $\log_b(a) - \log_b(c) = \log_b(\frac{a}{c})$ . Let's apply these properties step by step. First, simplify each term using the power rule: $3\log_b(2) = \log_b(2^3) = \log_b(8)$ , $\frac{1}{2}\log_b(9) = \log_b(9^{1/2}) = \log_b(3)$ , and $2\log_b(3) = \log_b(3^2) = \log_b(9)$ . Now the equation becomes: $\log_b(x) = \log_b(8) + \log_b(3) - \log_b(9)$ Using the addition and subtraction properties: $\log_b(x) = \log_b(\frac{8 \cdot 3}{9}) = \log_b(\frac{24}{9}) = \log_b(\frac{8}{3})$ Since the logarithms are equal, $x = \frac{8}{3}$ , which is choice D. Choice A ( $3$ ) likely comes from incorrectly simplifying $\log_b(9)^{1/2}$ . Choice B ( $\frac{24}{9}$ ) results from forgetting to reduce the fraction or misapplying the subtraction property. Choice C ( $8$ ) occurs when students ignore the other terms and focus only on $3\log_b(2) = \log_b(8)$ . Remember: always convert coefficients to exponents first, then combine logarithms using addition/subtraction rules. Work systematically through each property rather than trying shortcuts.
Which expression is equivalent to $2\log_6(x) - \log_6(9) + \log_6(3)$ when $x > 0$ ?
Answer: $\log_6(\frac{x^2}{3})$ When you encounter logarithmic expressions with multiple terms, your goal is to use logarithm properties to combine them into a single logarithm. The key properties you need are: $a\log_b(x) = \log_b(x^a)$ , $\log_b(x) + \log_b(y) = \log_b(xy)$ , and $\log_b(x) - \log_b(y) = \log_b(\frac{x}{y})$ . Let's work through $2\log_6(x) - \log_6(9) + \log_6(3)$ step by step. First, apply the power property: $2\log_6(x) = \log_6(x^2)$ . This gives us $\log_6(x^2) - \log_6(9) + \log_6(3)$ . Next, combine the addition and subtraction using the product and quotient properties. Since we have subtraction followed by addition, we get: $\log_6\left(\frac{x^2 \cdot 3}{9}\right)$ . Simplifying the fraction: $\frac{x^2 \cdot 3}{9} = \frac{3x^2}{9} = \frac{x^2}{3}$ . Therefore, the answer is $\log_6\left(\frac{x^2}{3}\right)$ , which is choice C. Choice A gives $\frac{x^2}{27}$ , which would result from incorrectly treating $+\log_6(3)$ as $-\log_6(3)$ . Choice B shows $\frac{x^2 \cdot 3}{9}$ without simplifying the fraction—this is the intermediate step, not the final answer. Choice D attempts to convert $\log_6(9)$ incorrectly, since $\log_6(9) \neq 2$ . Remember: always simplify fractions completely when working with logarithms, and double-check that you're applying addition and subtraction properties correctly—the order matters!
Which expression is equivalent to $\log_3(\frac{27x^2}{\sqrt{y}})$ when $x > 0$ and $y > 0$ ?
Answer: $3 + 2\log_3(x) - \frac{1}{2}\log_3(y)$ Using logarithm properties: $\log_3(\frac{27x^2}{\sqrt{y}}) = \log_3(27) + \log_3(x^2) - \log_3(\sqrt{y}) = \log_3(3^3) + 2\log_3(x) - \log_3(y^{1/2}) = 3 + 2\log_3(x) - \frac{1}{2}\log_3(y)$ . Choice B incorrectly applies the power rule to $\sqrt{y}$ as $2\log_3(y)$ . Choice C misapplies the quotient rule as division of expressions. Choice D fails to evaluate $\log_3(27)$ and doesn't apply power rules.
Which statement correctly justifies why $\log_3(x - 2) + \log_3(x + 2) = \log_3(x^2 - 4)$ is not valid for all real numbers $x$ ?
Answer: The equation requires $x - 2 > 0$ and $x + 2 > 0$ , so $x > 2$ When working with logarithmic equations, you must always consider the domain restrictions before applying algebraic properties. Logarithms are only defined for positive arguments, so each logarithmic term imposes its own constraint. For this equation to be valid, both $\log_3(x - 2)$ and $\log_3(x + 2)$ must be defined. This requires: - $x - 2 > 0$ , which means $x > 2$ - $x + 2 > 0$ , which means $x > -2$ Since both conditions must be satisfied simultaneously, you need $x > 2$ . Only when $x > 2$ can you apply the logarithm property $\log_a(m) + \log_a(n) = \log_a(mn)$ to get $\log_3(x^2 - 4)$ . This makes choice D correct. Choice A identifies the correct inequality $x^2 - 4 > 0$ (which gives $x < -2$ or $x > 2$ ), but this only ensures the right side is defined. The left side has stricter requirements that eliminate $x < -2$ . Choice B suggests $x > 0$ is sufficient, but this misses the crucial point that $x - 2$ must be positive, not just $x$ itself. Choice C mentions avoiding division by zero, but there's no division in this logarithmic equation. The issue is about keeping arguments positive, not avoiding zero denominators. Remember: when analyzing logarithmic equations, always check the domain of each individual logarithmic term first. The intersection of all domain restrictions determines where the equation is valid, regardless of what algebraic manipulations might suggest.

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