# 1967 AHSME Problems/Problem 4

## Problem

Given $\frac{\log{a}}{p}=\frac{\log{b}}{q}=\frac{\log{c}}{r}=\log{x}$, all logarithms to the same base and $x \not= 1$. If $\frac{b^2}{ac}=x^y$, then $y$ is:

$\text{(A)}\ \frac{q^2}{p+r}\qquad\text{(B)}\ \frac{p+r}{2q}\qquad\text{(C)}\ 2q-p-r\qquad\text{(D)}\ 2q-pr\qquad\text{(E)}\ q^2-pr$

## Solution

We are given: $$\frac{b^2}{ac} = x^y$$

Taking the logarithm on both sides: $$\log{\left(\frac{b^2}{ac}\right)} = \log{x^y}$$

Using the properties of logarithms: $$2\log{b} - \log{a} - \log{c} = y \log{x}$$

Substituting the values given in the problem statement: $$2q \log{x} - p \log{x} - r \log{x} = y \log{x}$$

Since $x \neq 1$, dividing each side by $\log{x}$ we get: $$y = \boxed{\textbf{(C) } 2q - p - r}$$

~ proloto

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