I need some help solving this inequality for a question involving the number of bounces, $n$, of ball such that the max. height of the ball is less than 5cm. This is the equation I have gathered from the previous information of the question.
$\frac{(\alpha^n\beta)^2}{2g}$ $<0.05$
I need $n>constant$ but I'm having difficulty with inequality signs with the logarithm.
Here $g=9.8ms^{-2}$, $\alpha<1$ and $\beta>1$.
Let $a,b,c,d>0$ and $n\in\Bbb N$. Then we have
$$ \frac{(a^nb)^2}{c}<d \iff (a^nb)^2<dc \iff a^nb<\sqrt{dc} \iff a^n < \frac{\sqrt{dc}}{b} \\ \iff n\ln(a)=\ln(a^n)<\ln\left(\frac{\sqrt{dc}}{b}\right) \qquad (*)$$
Where the fact that $x\mapsto \ln(x)$ is strictly increasing is used in the last equivalence.
Now if $a >1$, then $\ln(a)>0$ and $(*)$ is true if and only if $$n < \frac{\ln\left(\frac{\sqrt{dc}}{b}\right) }{\ln(a)}$$ and this shows that the constant you are looking for does not exists, i.e. there is no $C>0$ such that the inequality is satisfied for every $n>C$.
If $a=1$ then $\ln(a)=0$ and $(*)$ is true for every $n$ as long as $\frac{\sqrt{dc}}{b}>1$.
Finally if $a<1$, then $\ln(a)<0$ and $(*)$ is true if and only if $$n > \frac{\ln\left(\frac{\sqrt{dc}}{b}\right) }{\ln(a)}.$$