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

Now let $k = \log_{b} a$. Our equation becomes $$4k + \frac{3}{k} = 8$$ $$4k^2 - 8k + 3 = 0$$ $$k = \frac{8 \pm \sqrt{64-4(4)(3)}}{8}$$ $$k = \frac{1}{2}, \frac{3}{2}$$

Therefore, either $b = a^2$ or $b = \sqrt {a^2}$. Since no option matches both of these solutions, the answer is $\boxed{\textbf{(E)}}$.

- mako17