Solve each exponential equation:

Solution (a) We begin by expressing both sides of the equation with the same base so we can use the one-to-one property (1). $$\begin{array}{rcl@{\qquad}l} 4^{2x-1}&=&8^{x+3} \\[3pt] (2^{2}) ^{2x-1}&=& ( 2^{3}) ^{x+3} & {\color{#0066A7}{{\hbox{${4=2}^{2}, {8=2}^{3}$}}}} \\[3pt] 2^{2 ( 2x-1) }&=&2^{3( x+3) } & {\color{#0066A7}{{({a}^{r}) ^{s}{=a}^{rs}}}} \\[3pt] 2 ( 2x-1)~&=&3 ( x+3)~& {\color{#0066A7}{{\hbox{If } {a}^{u}=a^{v}, \hbox{then } u=v.}}} \\[3pt] 4x-2 &=& 3x+9 & {\color{#0066A7}{{\hbox{Simplify.}}}} \\[3pt] x &=& 11 & {\color{#0066A7}{{\hbox{Solve.}}}} \end{array}$$ The solution is 11.

(b) We use the Laws of Exponents to obtain the base $e$ on the right side. $$( e^{x}) ^{2}\cdot \dfrac{1}{e^{3}}=e^{2x}\cdot e^{-3}=e^{2x-3}$$ As a result, $$\begin{array}{rcl@{\qquad}l} e^{-x^{2}} &=&e^{2x-3} \\[3pt] -x^{2} &=&2x-3 & {\color{#0066A7}{{\hbox{If } {a}^{u}=a^{v}, \hbox{ then } u=v.}}} \\[3pt] x^{2}+2x-3 &=&0 \\[3pt] ( x+3) ( x-1)~&=&0 \\[3pt] x &=&-3\quad \hbox{or}\quad x=1 \end{array}$$ The solution set is $\{ -3,1\}$.