8.28 Let \(a, b\) and \(x\) be positive real numbers not equal to 1. Then, the equation
\[
4\left(\log _{a} x\right)^{2}+3\left(\log _{b} x\right)^{2}=8\left(\log _{a} x\right)\left(\log _{b} x\right)
\]

holds if and only if
(A) for all values of \(a, b\) and \(x\).
(B) if and only if \(a=b^{2}\).
(C) if and only if \(b=a^{2}\).
(D) if and only if \(x=a b\).
(E) none of these.
(27th American High School Mathematics Examination, 1976)
［Solution］The given equation can be written in the following form
$$
4\left(\log _{a} x\right)^{2}-8\left(\log _{a} x\right)\left(\log _{b} x\right)+3\left(\log _{b} x\right)^{2}=0 \text {, }
$$

which is
$$
\left(2 \log _{a} x-\log _{b} x\right)\left(2 \log _{a} x-3 \log _{b} x\right)=0 .
$$

Therefore, $\log _{a} x^{2}=\log _{b} x$, or $\log _{a} x^{2}=\log _{b} x^{3}$.
If we set $r=\log _{a} x^{2}$, then from the above equations we have
$a^{r}=x^{2}$ and $b^{r}=x$, or $a^{r}=x^{2}$ and $b^{r}=x^{3}$,
which means $a^{r}=b^{2 r}$, or $a^{3 r}=b^{2 r}$,
thus $a=b^{2}$ or $a^{3}=b^{2}$.
Therefore, the answer is $(E)$.