1976 AHSME Problems/Problem 20

Let $a,~b$ , and $x$ be positive real numbers distinct from one. Then $4(\log_ax)^2+3(\log_bx)^2=8(\log_ax)(\log_bx)$

$\textbf{(A) }\text{for all values of }a,~b,\text{ and }x\qquad\\ \textbf{(B) }\text{if and only if }a=b^2\qquad\\ \textbf{(C) }\text{if and only if }b=a^2\qquad\\ \textbf{(D) }\text{if and only if }x=ab\qquad\\ \textbf{(E) }\text{for none of these}$

Solution

Because $\log_{m} n = \dfrac{\log n}{\log m}$ , $4(\log_{a} x)^2+3(\log_{b} x)^2 =$ $\dfrac{4(\log x)^2}{(\log a)^2}+\dfrac{3(\log x)^2}{(\log b)^2} = \dfrac{(\log x)^2(4(\log a)^2+3(\log b)^2)}{(\log a \log b)^2}$ .

Therefore,

$\dfrac{(\log x)^2(4(\log a)^2+3(\log b)^2)}{(\log a \log b)^2} = \frac{8(\log x)^2}{\log a \log b}$ $\dfrac{4(\log a)^2+3(\log b)^2}{\log a \log b} = 8$