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

1967 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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1967 AHSME Problems/Problem 4

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