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

m
m (Solution 2)
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that then becomes <math>\log_2 (4)</math> * <math>\log_{3x} (3x)</math> = <math>\log_2 (8)</math> * <math>\log_{2x} (3x)</math>
 
that then becomes <math>\log_2 (4)</math> * <math>\log_{3x} (3x)</math> = <math>\log_2 (8)</math> * <math>\log_{2x} (3x)</math>
  
which simplifies to 2*1 = 3<math>\log_{2x} (3x)</math>
+
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
 
so now <math>\frac{2}{3}</math> = <math>\log_{2x} (3x)</math> putting in exponent form gets
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<math>(2x)^2</math> = <math>(3x)^3</math>
 
<math>(2x)^2</math> = <math>(3x)^3</math>
  
so 4<math>x^2</math> = 27<math>x^3</math>
+
so <math>4x^2</math> = <math>27x^3</math>
  
dividing yields x = 4/27 and  
+
dividing yields <math>x = 4/27</math> and  
  
4+27 = <math>\boxed{\textbf{(D)}31}</math>
+
<math>4+27 =</math> <math>\boxed{\textbf{(D)}31}</math>
 
  - Pikachu13307
 
  - Pikachu13307
  

Revision as of 03:55, 25 January 2019

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

See Also

2018 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AMC 12 Problems and Solutions

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