# Difference between revisions of "1964 AHSME Problems/Problem 21"

## Problem 21

If $\log_{b^2}x+\log_{x^2}b=1, b>0, b \neq 1, x \neq 1$, then $x$ equals: $\textbf{(A)}\ 1/b^2 \qquad \textbf{(B)}\ 1/b \qquad \textbf{(C)}\ b^2 \qquad \textbf{(D)}\ b \qquad \textbf{(E)}\ \sqrt{b}$

## Solution

Using natural log as a "neutral base", and applying the change of base formula to each term, we get: $\frac{\ln x}{\ln b^2} + \frac{\ln b}{\ln x^2} = 1$ $\frac{\ln x}{2\ln b} + \frac{\ln b}{2\ln x} = 1$ $\frac{\ln x \ln x + \ln b \ln b}{2\ln b \ln x} = 1$ $\ln x \ln x + \ln b \ln b = 2\ln b \ln x$

You could inspect the equation here and see that $x=b$ is one solution. Or, you can substitute $X = \ln x$ and $B = \ln b$ to get a quadratic in $X$: $X^2 + B^2 = 2BX$ $X^2 - 2BX + B^2 = 0$

The above is a quadratic with coefficients $(1, -2B, B^2)$. Plug into the QF to get: $X = \frac{2B \pm \sqrt{4B^2 - 4B^2}}{2}$ $X = B$ $\ln x = \ln b$ $x = b$

Either way, the answer is $\boxed{\textbf{(D)}}$.

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