Answer
(a) $-3$
(b) $\frac{\sqrt{2}}{2}$
(c) $\frac{1}{64}$
Work Step by Step
(a) $\ln \frac{1}{e^3}=\ln e^{-3}=-3\ln e=-3\cdot 1=-3$
(b)
Let $x=\tan^{-1} 1$.
We have $\tan (x)=1$ and since $x\in (-\pi/2,\pi/2)$, $x=\pi/4$.
So, $\sin (\tan^{-1}1)=\sin x=\sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}$.
(c) $10^{-3\log 4}=10^{(\log 4)\cdot (-3)}=(10^{\log 4})^{-3}=4^{-3}=\frac{1}{4^3}=\frac{1}{64}$
