Calculus: Early Transcendentals, Chapter 1, 1.6, Section 1.6, Problem 50

Evaluate the equation (a) $\ln (\ln x) = 1$ (b) $e^{ax} = Ce^{bx}$, where $a \neq b$ for $x$

a.) $\ln(\ln x) = 1$

$
\begin{equation}
\begin{aligned}
e^{\ln (\ln x)} &= e^1
&& \text{Raise in $e$ both sides to eliminate the first } \ln\\
\\
\ln x &= e^1
&& \text{Evaluate}\\
\\
e^{\ln x} &= e^{e^1}
&& \text{Repeat the first part and solve for } x\\
\\
x &= e^{e^1}
\end{aligned}
\end{equation}
$


b.) $e^{ax} = Ce^{bx}$

$
\begin{equation}
\begin{aligned}
\frac{e^{ax}}{e^{bx}} &= C
&& \text{Divide each side by } e^{bx}\\
\\
e^{ax - bx} &= C
&& \text{Apply the property of exponent}\\
\\
e^{x(a -b)} &= C
&& \text{Factor $x$ in the exponent}\\
\\
\ln e^{x(a - b)} &= \ln C
&& \text{Take ln of each side}\\
\\
x (a - b) &= \ln C
&& \text{Simplify}\\
\\
x &= \frac{\ln C}{a - b}
&& \text{Solve for } x

\end{aligned}
\end{equation}
$