Difference between revisions of "2018 AMC 12A Problems/Problem 14"

(Solution 1)
(Polished Sol 2.)
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==Solution 2==
 
==Solution 2==
If you multiply both sides by <math>\log_2 (3x)</math>
+
By the logarithmic identity <math>n\log_b{a}=\log_b{\left(a^n\right)},</math> the original equation becomes <cmath>2\log_{3x} 2 = 3\log_{2x} 2.</cmath>
 +
By the logarithmic identity <math>\log_b{a}\cdot\log_a{b}=1,</math> we multiply both sides by <math>\log_2{(2x)},</math> then apply the Change of Base Formula to the left side:
 +
<cmath>\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*}</cmath>
 +
Therefore, the answer is <math>4+27=\boxed{\textbf{(D) } 31}.</math>
  
then it should come out to <math>\log_2 (3x)</math> * <math>\log_{3x} (4)</math> = <math>\log_2 {3x}</math> * <math>\log_{2x} (8)</math>
+
~Pikachu13307 (Fundamental Logic)
  
that then becomes <math>\log_2 (4)</math> * <math>\log_{3x} (3x)</math> = <math>\log_2 (8)</math> * <math>\log_{2x} (3x)</math>
+
~MRENTHUSIASM (Reconstruction)
 
 
which simplifies to <math>2*1 = 3\log_{2x} (3x)</math>
 
 
 
so now <math>\frac{2}{3}</math> = <math>\log_{2x} (3x)</math> putting in exponent form gets
 
 
 
<math>(2x)^2</math> = <math>(3x)^3</math>
 
 
 
so <math>4x^2</math> = <math>27x^3</math>
 
 
 
dividing yields <math>x = 4/27</math> and
 
 
 
<math>4+27 =</math> <math>\boxed{\textbf{(D)}31}</math>
 
- Pikachu13307
 
  
 
==Solution 3==
 
==Solution 3==

Revision as of 08:31, 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 3

We can convert both $4$ and $8$ into $2^2$ and $2^3$, respectively, giving:

$2\log_{3x} (2) = 3\log_{2x} (2)$

Converting the bases of the right side, we get $\log_{2x} 2 = \frac{\ln 2}{\ln (2x)}$

$\frac{2}{3}*\log_{3x} (2) = \frac{\ln 2}{\ln (2x)}$

$2^\frac{2}{3} = (3x)^\frac{\ln 2}{\ln (2x)}$

$\frac{2}{3} * \ln 2 = \frac{\ln 2}{\ln (2x)} * \ln (3x)$

Dividing both sides by $\ln 2$, we get

$\frac{2}{3} = \frac{\ln (3x)}{\ln (2x)}$

Which simplifies to

$2\ln (2x) = 3\ln (3x)$

Log expansion allows us to see that

$2\ln 2 + 2\ln (x) = 3\ln 3 + 3\ln (x)$, which then simplifies to

$\ln (x) = 2\ln 2 - 3\ln 3$

Thus,

$x = e^{2\ln 2 - 3\ln 3} = \frac{e^{2\ln 2}}{e^{3\ln 3}}$

And

$x = \frac{2^2}{3^3} = \frac{4}{27} = \boxed{\textbf{(D)}31}$

-lepetitmoulin

Solution 4

$\log_{3x} 4=\log_{2x} 8$ 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 5

$\log_{3x} 4=\log_{2x} 8\implies 2\log_{3x} 2=3\log_{2x} 2 \implies \frac{2}{3}=\frac{\log_{2x}2}{\log_{2x}3}$. 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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