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# Logarithm and Equations | AIME I, 2012 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2015 based on Logarithm and Equations.

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on logarithm and Equations.

## Logarithm and Equations – AIME I, 2012

Let x,y,z be positive real numbers $2log_{x}(2y)$=$2log_{2x}(4z)=log_{2x^4}(8yz)\neq0$ the value of (x)($y^{5}$)(z) may be expressed in the form $\frac{1}{2^\frac{p}{q}}$ where p and q are relatively prime positive integers, find p+q.

• is 107
• is 49
• is 840
• cannot be determined from the given information

### Key Concepts

Equations

Algebra

Logarithm

AIME I, 2012, Question 9

Higher Algebra by Hall and Knight

## Try with Hints

Let $2log_{x}(2y)$=$2log_{2x}(4z)=log_{2x^4}(8yz) =2$ then from first and last term x=2y from second and last term 2x=4z and from third and last term $4x^{8}=8yz$

taking these together $4x^{8}$=(4z)(2y)=x(2x) then x=$2^\frac{-1}{6}$ then y=z=$2^\frac{-7}{6}$

(x)($y^{5}$)(z) =$2^\frac{-43}{6}$ then p+q =43+6=49.