2018 AMC 12A Problems/Problem 14

Revision as of 09:18, 14 August 2021 by MRENTHUSIASM (talk | contribs) (Polished Sol 5.)

Problem

The solutions to the equation $\log_{3x} 4 = \log_{2x} 8$, where $x$ is a positive real number other than $\tfrac{1}{3}$ or $\tfrac{1}{2}$, can be written as $\tfrac {p}{q}$ where $p$ and $q$ are relatively prime positive integers. What is $p + q$?

$\textbf{(A) } 5   \qquad     \textbf{(B) } 13   \qquad     \textbf{(C) } 17   \qquad    \textbf{(D) } 31 \qquad   \textbf{(E) } 35$

Solution 1

We apply the Change of Base Formula, then rearrange: \begin{align*} \frac{\log_2{4}}{\log_2{(3x)}}&=\frac{\log_2{8}}{\log_2{(2x)}} \\ \frac{2}{\log_2{(3x)}}&=\frac{3}{\log_2{(2x)}} \\ 3\log_2{(3x)}&=2\log_2{(2x)}. \\ \end{align*} By the logarithmic identity $n\log_b{a}=\log_b{\left(a^n\right)},$ it follows that \begin{align*} \log_2{\left[(3x)^3\right]}&=\log_2{\left[(2x)^2\right]} \\ (3x)^3&=(2x)^2\\ 27x^3&=4x^2 \\ x&=\frac{4}{27}, \end{align*} from which the answer is $4+27=\boxed{\textbf{(D) } 31}.$

~jeremylu (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Solution 2

By the logarithmic identity $n\log_b{a}=\log_b{\left(a^n\right)},$ the original equation becomes \[2\log_{3x} 2 = 3\log_{2x} 2.\] By the logarithmic identity $\log_b{a}\cdot\log_a{b}=1,$ we multiply both sides by $\log_2{(2x)},$ then apply the Change of Base Formula to the left side: \begin{align*} 2\left[\log_{3x}2\right]\left[\log_2{(2x)}\right] &= 3 \\ 2\left[\frac{\log_2 2}{\log_2{(3x)}}\right]\left[\frac{\log_2{(2x)}}{\log_2 2}\right] &= 3 \\ 2\left[\frac{\log_2{(2x)}}{\log_2{(3x)}}\right] &=3 \\ 2\left[\log_{3x}{(2x)}\right] &= 3 \\ \log_{3x}{\left[(2x)^2\right]} &= 3 \\ (3x)^3&=(2x)^2\\ 27x^3&=4x^2 \\ x&=\frac{4}{27}. \end{align*} Therefore, the answer is $4+27=\boxed{\textbf{(D) } 31}.$

~Pikachu13307 (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Solution 4

We can convert both $4$ and $8$ into $2^2$ and $2^3,$ respectively: \[2\log_{3x} (2) = 3\log_{2x} (2).\] Converting the bases of the right side, we get \begin{align*} \log_{2x} 2 &= \frac{\ln 2}{\ln (2x)} \\ \frac{2}{3}\cdot\log_{3x} (2) &= \frac{\ln 2}{\ln (2x)} \\ 2^\frac{2}{3} &= (3x)^\frac{\ln 2}{\ln (2x)} \\ \frac{2}{3} \cdot \ln 2 &= \frac{\ln 2}{\ln (2x)} \cdot \ln (3x). \end{align*} Dividing both sides by $\ln 2,$ we get $\frac{2}{3} = \frac{\ln (3x)}{\ln (2x)},$ from which \[2\ln (2x) = 3\ln (3x).\] Expanding this equation gives \begin{align*} 2\ln 2 + 2\ln (x) &= 3\ln 3 + 3\ln (x) \\ \ln (x) &= 2\ln 2 - 3\ln 3. \end{align*} Thus, we have \[x = e^{2\ln 2 - 3\ln 3} = \frac{e^{2\ln 2}}{e^{3\ln 3}} = \frac{2^2}{3^3} = \frac{4}{27},\] from which the answer is $4+27=\boxed{\textbf{(D) } 31}.$

~lepetitmoulin (Solution)

~MRENTHUSIASM (Reformatting)

Solution 5

Note that $\log_{3x} 4=\log_{2x} 8$ is the same as \[2\log_{3x} 2=3\log_{2x} 2.\] Using Reciprocal law, we get \begin{align*} \log_{(3x)^\frac{1}{2}} 2&=\log_{(2x)^\frac{1}{3}} 2 \\ (3x)^\frac{1}{2}&=(2x)^\frac{1}{3} \\ 27x^3&=4x^2 \\ x&=\frac{4}{27}, \end{align*} from which the answer is $4+27=\boxed{\textbf{(D) } 31}.$

~OlutosinNGA (Solution)

~MRENTHUSIASM (Reformatting)

Solution 6

$\log_{3x} 4=\log_{2x} 8\implies 2\log_{3x} 2=3\log_{2x} 2 \implies \frac{2}{3}=\frac{\log_{2x}2}{\log_{2x}3}.$ We know that $\log_a{b}=\frac{\log_{c}b}{\log_{c}a}=\frac{\frac{1}{\log_b{c}}}{\frac{1}{\log_a{c}}}=\frac{\log_a{c}}{\log_b{c}}.$ Thus $\frac{2}{3}=\frac{\log_{2x}2}{\log_{2x}3}\implies \frac{2}{3}=\log_{2x}{3x}\implies (2x)^{\frac{2}{3}}=3x\implies 2^{\frac{2}{3}}x^{\frac{2}{3}}=3x\implies 2^{\frac{2}{3}}=3x^{\frac{1}{3}}\implies x^{\frac{1}{3}}=\frac{2^{\frac{2}{3}}}{3}\implies x=\frac{2^2}{3^3}=\frac{4}{27}.$ $4$ and $27$ are indeed relatively prime thus our final answer is $4+27=31 \text{which is }\boxed{\textbf{(D)}}$

-vsamc

See Also

2018 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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