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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

Check the Answer


Answer: is 49.

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.

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