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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

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 $y_{1}+y_{2}=25$.

Categories

## 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

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.