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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

Equations

Algebra

Logarithm

## Check the Answer

But try the problem first…

Source

AIME I, 2012, Question 9

Higher Algebra by Hall and Knight

## Try with Hints

First hint

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 $4x^{8}=8yz$

Second Hint

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

Final Step

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