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2018 AMC 12A Problems/Problem 14: Difference between revisions

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==Problem==
==Problem==


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


<math>\textbf{(A) } 5  \qquad     
<math>\textbf{(A) } 5  \qquad     
Line 85: Line 85:


==Solution 5 (Exponential Form)==
==Solution 5 (Exponential Form)==
Let <math>y=\log_{3x} 4 = \log_{2x} 8.</math> We convert each equation for <math>y</math> to the exponential form:
Let <math>y=\log_{3x} 4 = \log_{2x} 8.</math> We convert the equations with <math>y</math> to the exponential form:
<cmath>\begin{align*}
<cmath>\begin{align*}
(3x)^y&=4, \\
(3x)^y&=4, \\

Latest revision as of 11:37, 23 July 2024

Problem

The solution to the equation $\log_{3x} 4 = \log_{2x} 8$, where $x$ is a positive real number other than $\frac{1}{3}$ or $\frac{1}{2}$, can be written as $\frac {p}{q}$ 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

We will apply the following logarithmic identity: \[\log_{p^n}{\left(q^n\right)}=\log_{p}{q},\] which can be proven by the Change of Base Formula: \[\log_{p^n}{\left(q^n\right)}=\frac{\log_{p}{\left(q^n\right)}}{\log_{p}{\left(p^n\right)}}=\frac{n\log_{p}{q}}{n}=\log_{p}{q}.\] We rewrite the original equation as $\log_{(3x)^3} 64 = \log_{(2x)^2} 64,$ from which \begin{align*} (3x)^3&=(2x)^2 \\ 27x^3&=4x^2 \\ x&=\frac{4}{27}. \end{align*} Therefore, the answer is $4+27=\boxed{\textbf{(D) } 31}.$

~MRENTHUSIASM

Solution 3

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 (Exponential Form)

Let $y=\log_{3x} 4 = \log_{2x} 8.$ We convert the equations with $y$ to the exponential form: \begin{align*} (3x)^y&=4, \\ (2x)^y&=8. \end{align*} Cubing the first equation and squaring the second equation, we have \begin{align*} (3x)^{3y}&=64, \\ (2x)^{2y}&=64. \end{align*} Applying the Transitive Property, we get \begin{align*} (3x)^{3y}&=(2x)^{2y} \\ (3x)^3&=(2x)^2 \\ 27x^3&=4x^2 \\ x&=\frac{4}{27}, \end{align*} from which the answer is $4+27=\boxed{\textbf{(D) } 31}.$

~MRENTHUSIASM

See Also

2018 AMC 12A (Problems • Answer Key • Resources)
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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