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

## Check the Answer

But try the problem first…

Answer: is 49.

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.

## Other useful links

- https://www.cheenta.com/cubes-and-rectangles-math-olympiad-hanoi-2018/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s

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