A Mistake I Made

A counter-example: $(\frac{1}{2})^2 = \frac{1}{4}$ and $(\frac{1}{2})^3 = \frac{1}{8}$ .

Now, $\frac{1}{8} < \frac{1}{4}$ but $\log_{\frac{1}{2}} {\frac{1}{8}} = 3 \not< \log_{\frac{1}{2}}{\frac{1}{4}} = 2$ .

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Generally, the statement in the problem description hold only when $b >1$ ; even when any one or both of $p$ and $q$ can be $1$ .

And the $<$ sign in the statement in the problem description gets replaced by $>$ , making the statement true, when $b <1$ ; even when any one or both of $p$ and $q$ can be $1$ .