Logarithmic Equation Solution Count

Simplify using change of base formula: $\frac{\log_a(16)}{\log_a(32)} = \log_{32}(16) = \frac{4}{5}$ $\frac{\log_a(64)}{\log_a(32)} = \log_{32}(64) = \frac{6}{5}$

The equation becomes: $\frac{4}{5} x^2 - x - \frac{6}{5} = 0 \implies 4x^2 - 5x - 6 = 0$. Discriminant $D = 25 - 4(4)(-6) = 121 > 0$, giving 2 real roots. However, the logarithmic terms $\log_a(16), \log_a(32)$ require base $a > 0$ and $a \neq 1$. If $a = 1$, the logarithm is undefined. Therefore, the existence of solutions depends on whether $a \neq 1$, i.e., depends on the value of $a$.