Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on logarithm and Equations.
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.
Equations
Algebra
Logarithm
Answer: is 49.
AIME I, 2012, Question 9
Higher Algebra by Hall and Knight
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.