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Logarithms and Equations | AIME I, 2000 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 2000 based on Logarithms and Equations.

Logarithms and Equations – AIME I 2000


\(log_{10}(2000xy)-log_{10}xlog_{10}y=4\) and \(log_{10}(2yz)-(log_{10}y)(log_{10}z)=1\) and \(log_{10}(zx)-(log_{10}z)(log_{10}x)=0\) has two solutions \((x_{1},y_{1},z_{1}) and (x_{2},y_{2},z_{2})\) find \(y_{1}+y_{2}\).

  • is 905
  • is 25
  • is 840
  • cannot be determined from the given information

Key Concepts


Logarithms

Theory of Equations

Number Theory

Check the Answer


Answer: is 25.

AIME I, 2000, Question 9

Polynomials by Barbeau

Try with Hints


Rearranging equations we get \(-logxlogy+logx+logy-1=3-log2000\) and \(-logylogz+logy+logz-1=-log2\) and \(-logxlogz+logx+logz-1=-1\)

taking p, q, r as logx, logy and logz, \((p-1)(q-1)=log2\) and \((q-1)(r-1)=log2\) and \( (p-1)(r-1)=1\) which is first system of equations and multiplying the first three equations of the first system gives \((p-1)^{2}(q-1)^{2}(r-1)^{2}=(log 2)^{2}\) gives \((p-1)(q-1)(r-1)=+-(log2)\) which is second equation

from both equations (q-1)=+-(log2) gives (logy)+-(log2)=1 gives \(y_{1}=20\),\(y_{2}=5\) then \(y_{1}+y_{2}=25\).

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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, 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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