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

Revision as of 18:20, 5 April 2020

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

Base switch to log 2 and you have $\frac{\log_2 4}{\log_2 3x} = \frac{\log_2 8}{\log_2 2x}$ .

$\frac{2}{\log_2 3x} = \frac{3}{\log_2 2x}$

$2*\log_2 2x = 3*\log_2 3x$

Then $\log_2 (2x)^2 = \log_2 (3x)^3$ . so $4x^2=27x^3$ and we have $x=\frac{4}{27}$ leading to $\boxed{\textbf{(D)}31}$ (jeremylu)

Solution 2

If you multiply both sides by $\log_2 (3x)$

then it should come out to $\log_2 (3x)$ * $\log_{3x} (4)$ = $\log_2 {3x}$ * $\log_{2x} (8)$

that then becomes $\log_2 (4)$ * $\log_{3x} (3x)$ = $\log_2 (8)$ * $\log_{2x} (3x)$

which simplifies to $2*1 = 3\log_{2x} (3x)$

so now $\frac{2}{3}$ = $\log_{2x} (3x)$ putting in exponent form gets

$(2x)^2$ = $(3x)^3$

so $4x^2$ = $27x^3$

dividing yields $x = 4/27$ and

$4+27 =$ $\boxed{\textbf{(D)}31}$

- Pikachu13307

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

@@ Line 88: / Line 88: @@
 ~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>
+-vsamc
 ==See Also==
 {{AMC12 box|year=2018|ab=A|num-b=13|num-a=15}}
 {{MAA Notice}}

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