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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 $\boxed{b}$.