7. For what value of $a>1$ is
$$
\int_{a}^{a^{2}} \frac{1}{x} \log \frac{x-1}{32} d x
$$
minimum?
Solution:
3
Let $f(a)=\int_{a}^{a^{2}} \frac{1}{x} \log \frac{x-1}{32} d x$. Then we want $\frac{d f}{d a}=0$; by the Fundamental Theorem of Calculus and the chain rule, this implies that
$$
2 a\left(\frac{1}{a^{2}} \log \frac{a^{2}-1}{32}\right)-\frac{1}{a} \log \frac{a-1}{32}=\frac{d}{d a}\left(\int_{c}^{a^{2}} \frac{1}{x} \log \frac{x-1}{32} d x-\int_{c}^{a} \frac{1}{x} \log \frac{x-1}{32} d x\right)=0,
$$
where $c$ is any constant with $11$ ), this simplifies to $\left(a^{2}-1\right)(a+1)=$ $32 \Rightarrow a^{3}+a^{2}-a-33=0$, which in turn factors as $(a-3)\left(a^{2}+4 a+11\right)=0$. The quadratic factor has no real solutions, so this leaves only $a=3$. However, we have that $a>1$, and we can check that $f(1)=0, \lim _{a \rightarrow \infty} f(a)>0$, and $f(3)<0$, so the global minimum does occur at $a=3$.