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

(Created page with "== 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 <ma...")
 
(→‎Solution)
Line 20: Line 20:
 
Then <math>\log_2 (2x)^2 = \log_2 (3x)^3</math>. so <math>4x^2=27x^3</math> and we have <math>x=\frac{4}{27}</math> leading to <math>\boxed{ (D) 31}</math> (jeremylu)
 
Then <math>\log_2 (2x)^2 = \log_2 (3x)^3</math>. so <math>4x^2=27x^3</math> and we have <math>x=\frac{4}{27}</math> leading to <math>\boxed{ (D) 31}</math> (jeremylu)
  
 +
==Solution 2==
 +
If you multiply both sides by <math>\log_2 (3x)</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>
 +
 +
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>
 +
 +
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 4<math>x^2</math> = 27<math>x^3</math>
 +
 +
dividing yields x = 4/27 and
 +
 +
4+27 = <math>\boxed{ (D) 31}</math>
 +
- Pikachu13307
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2018|ab=A|num-b=13|num-a=15}}
 
{{AMC12 box|year=2018|ab=A|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:56, 9 February 2018

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

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

dividing yields x = 4/27 and

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

- Pikachu13307

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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