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. The three logarithm properties come from exponent rules and work because logarithms are exponents: (1) Product property: ln(xy) = ln(x) + ln(y), (2) Quotient property: ln(x/y) = ln(x) - ln(y), (3) Power property: $ln(x^p$) = p ln(x). To expand ln( $(a^3$ sqrt(b)) / c ), first quotient: $ln(a^3$ sqrt(b)) - ln(c), then product: $ln(a^3$) + ln(sqrt(b)) - ln(c), and power: 3 ln(a) + (1/2) ln(b) - ln(c). Choice B correctly applies the properties to fully expand with coefficients. Choice D incorrectly keeps $a^3$ + $b^{1/2}$ inside a log—there's no sum property, it's for products! Expand by addressing quotients first, then products, then powers, working outside in. You're doing wonderfully—keep applying this step-by-step!

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. 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. To expand log_$3((x^2$ $y)/z^3$), first apply the quotient property: log_$3(x^2$ y) - log_$3(z^3$), then apply the product property to the numerator: log_$3(x^2$) + log_3(y) - log_$3(z^3$), and finally the power property: 2 log_3(x) + log_3(y) - 3 log_3(z). Choice B correctly applies these properties in sequence to fully expand the expression into a sum and difference of simpler logs with coefficients. Choice D incorrectly combines $x^2$ + y inside a log instead of using the product property for $x^2$ y, remember there's no property for log of a sum—it's a common mix-up with the product rule! When expanding logarithms, start with the outermost operation like quotients or overall powers, then break down products and individual powers step by step. This systematic approach ensures you don't miss any terms—keep practicing, you've got this!

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 with coefficients. 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/y) = log_b(x) - log_b(y), (3) Power property: log_$b(x^p$) = p log_b(x). To expand log_7( (x $y^2$$)^3$ / sqrt(z) ), first quotient: log_7( (x $y^2$$)^3$ ) - log_7( sqrt(z) ), then power: 3 log_7(x $y^2$) - (1/2) log_7(z), then product and power again: 3 (log_7(x) + 2 log_7(y)) - (1/2) log_7(z) = 3 log_7(x) + 6 log_7(y) - (1/2) log_7(z). Choice B correctly expands by applying properties from outside in, handling the overall power first. Choice C forgets to multiply the inner coefficients by 3—always distribute outer powers fully! Break down by identifying outermost structures like powers or quotients first, then recurse inward. You're mastering this—keep going!

This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you condense sums and differences into a single logarithm. The three logarithm properties come from exponent rules and work because logarithms are exponents: (1) Product property: ln(x) + ln(y) = ln(xy), (2) Quotient property: ln(x) - ln(y) = ln(x/y), (3) Power property: p ln(x) = $ln(x^p$). To condense (3/2) ln(x) + ln(y) - (1/2) ln(z), first power: $ln(x^{3/2}$) + ln(y) - $ln(z^{1/2}$), then product: $ln(x^{3/2}$ y) - $ln(z^{1/2}$), then quotient: ln( $(x^{3/2}$ y) / $z^{1/2}$ ). Choice A correctly applies the reverse properties to form the condensed expression. Choice B treats coefficients as multipliers inside instead of exponents—power rule pulls them up! Handle fractional coefficients with power first, then combine step by step. You're getting the hang of this—superb!

This question tests your understanding of logarithm properties in the context of pH calculation, using the power property to simplify. The power property is $log_b(x^p) = p log_b(x)$, but here we use the definition: $log(10^{-7}) = -7$, since $log(10^k) = k$ for base 10. For pH = $- log(10^{-7})$, that's $- (-7) = 7$, directly from the power simplifying inside the log. Choice B correctly applies the property to find the positive value after the negative sign. Choice A forgets the outer negative sign in pH definition—always include it! Recognize when arguments are powers of the base, simplify inside first, then apply outer operations. This makes applied problems like pH straightforward—you're doing fantastically!

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!

This question tests your understanding of the change-of-base formula, which is derived from logarithm properties, to rewrite logs in terms of natural logs. The change-of-base formula is log_b(x) = ln(x) / ln(b), coming from the power property and definition of logs, allowing conversion between bases. To express log_5(20) in natural logs, directly apply the formula: ln(20) / ln(5). Choice B correctly uses the change-of-base formula with the argument in the numerator and base in the denominator. Choice A swaps numerator and denominator—remember, it's log of x over log of b! Use change-of-base when the base doesn't match known values, always putting the argument's log on top. This tool opens up any log calculation—keep up the great work!

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!

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!

This question tests expanding complex logs with properties. Properties simplify nested expressions! To expand $ \log_2 \left( \frac{(x y^2)^3}{\sqrt{z}} \right) $: (1) Quotient: $ \log_2( (x y^2)^3 ) - \log_2( z^{1/2} ) $. (2) Power: $ 3 \log_2( x y^2 ) - \frac{1}{2} \log_2(z) $. (3) Product and power inside: $ 3 (\log_2(x) + 2 \log_2(y)) - \frac{1}{2} \log_2(z) = 3 \log_2(x) + 6 \log_2(y) - \frac{1}{2} \log_2(z) $. Choice A correctly expands. Choice B uses 2 instead of 6 for y—remember to distribute the 3 to y^2's log! Strategy: (1) Outermost (quotient). (2) Outer power. (3) Inner product/power. Work inside out—you're excelling!