# 1964 AHSME Problems/Problem 21

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## 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 $\boxed{\textbf{(D)}}$.