How to solve:
Hello, thebristolsound!
Solve: .$\displaystyle e^{2x} + 3e^x - 10 \:=\:0$
Factor: .$\displaystyle (e^x + 5)(e^x - 2) \:=\:0$
Set each factor equal to zero and solve.
. . $\displaystyle e^x + 5 \:=\:0 \quad\Rightarrow\quad e^x \:=\:-5$ . . . no real roots
. . $\displaystyle e^x - 2 \:=\:0 \quad\Rightarrow\quad e^x \:=\:2 \quad\Rightarrow\quad\boxed{ x \:=\:\ln 2}$