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

Revision as of 09:10, 14 August 2021

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

$\log_{3x} 4=\log_{2x} 8$ is the same as $2\log_{3x} 2=3\log_{2x} 2$

Using Reciprocal law, we get $\log_{(3x)^\frac{1}{2}} 2=\log_{(2x)^\frac{1}{3}} 2$

$\Rightarrow (3x)^\frac{1}{2}=(2x)^\frac{1}{3}$ $\Rightarrow 27x^3=4x^2$ $\Rightarrow \frac{x^3}{x^2}=\frac{4}{27}=x$

$\therefore \frac{p}{q}=\frac{4}{27}$ $\Rightarrow p+q=4+27=$ $\boxed{\textbf{(D) } 31}$

~OlutosinNGA

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

@@ Line 48: / Line 48: @@
 ~MRENTHUSIASM (Reconstruction)
-==Solution 3==
+==Solution 4==
+We can convert both <math>4</math> and <math>8</math> into <math>2^2</math> and <math>2^3,</math> respectively: <cmath>2\log_{3x} (2) = 3\log_{2x} (2).</cmath>
+Converting the bases of the right side, we get
+<cmath>\begin{align*}
+\log_{2x} 2 &= \frac{\ln 2}{\ln (2x)} \\
+\frac{2}{3}\cdot\log_{3x} (2) &= \frac{\ln 2}{\ln (2x)} \\
+^\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*}</cmath>
+Dividing both sides by <math>\ln 2,</math> we get <math>\frac{2}{3} = \frac{\ln (3x)}{\ln (2x)},</math> from which <cmath>2\ln (2x) = 3\ln (3x).</cmath>
+Expanding this equation gives
+<cmath>\begin{align*}
+\ln 2 + 2\ln (x) &= 3\ln 3 + 3\ln (x) \\
+\ln (x) &= 2\ln 2 - 3\ln 3.
+\end{align*}</cmath>
+Thus, we have <cmath>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},</cmath>
+from which the answer is <math>4+27=\boxed{\textbf{(D) } 31}.</math>
-We can convert both <math>4</math> and <math>8</math> into <math>2^2</math> and <math>2^3</math>, respectively, giving:
+~lepetitmoulin (Solution)
-<math>2\log_{3x} (2) = 3\log_{2x} (2)</math>
+~MRENTHUSIASM (Reformatting)
-Converting the bases of the right side, we get <math>\log_{2x} 2 = \frac{\ln 2}{\ln (2x)}</math>
+==Solution 5==
-<math>\frac{2}{3}*\log_{3x} (2) = \frac{\ln 2}{\ln (2x)}</math>
-<math>2^\frac{2}{3} = (3x)^\frac{\ln 2}{\ln (2x)}</math>
-<math>\frac{2}{3} * \ln 2 = \frac{\ln 2}{\ln (2x)} * \ln (3x)</math>
-Dividing both sides by <math>\ln 2</math>, we get
-<math>\frac{2}{3} = \frac{\ln (3x)}{\ln (2x)}</math>
-Which simplifies to
-<math>2\ln (2x) = 3\ln (3x)</math>
-Log expansion allows us to see that
-<math>2\ln 2 + 2\ln (x) = 3\ln 3 + 3\ln (x)</math>, which then simplifies to
-<math>\ln (x) = 2\ln 2 - 3\ln 3</math>
-Thus,
-<math>x = e^{2\ln 2 - 3\ln 3} = \frac{e^{2\ln 2}}{e^{3\ln 3}}</math>
-And
-<math>x = \frac{2^2}{3^3} = \frac{4}{27} = \boxed{\textbf{(D)}31}</math>
--lepetitmoulin
-==Solution 4==
 <math>\log_{3x} 4=\log_{2x} 8</math> is the same as <math>2\log_{3x} 2=3\log_{2x} 2</math>
@@ Line 96: / Line 80: @@
 ~OlutosinNGA
-==Solution 5==
-<math>\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}</math>. We know that <math>\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}}</math>. Thus <math>\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}</math>. <math>4</math> and <math>27</math> are indeed relatively prime thus our final answer is <math>4+27=31 \text{which is }\boxed{\textbf{(D)}}</math>
+==Solution 6==
+<math>\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}.</math> We know that <math>\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}}.</math> Thus <math>\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}.</math> <math>4</math> and <math>27</math> are indeed relatively prime thus our final answer is <math>4+27=31 \text{which is }\boxed{\textbf{(D)}}</math>
 -vsamc

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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