Solving exponential equations with logarithms (ACT). Game on.

Divide both sides by $2$ to get $e^{x}=7$. Taking $\ln$ of both sides gives $x=\ln 7\textsf{.}$

Take $\ln$ of both sides: $x+3=\ln 8$. Subtract $3$ to get $x=\ln 8-3\textsf{.}$

Take $\log$ of both sides: $x\log 4=\log 20$. Divide by $\log 4$ to get $x=\dfrac{\log 20}{\log 4}\textsf{.}$

Take $\ln$ of both sides: $-x=\ln 6$. Since $-x=(-1)x$, divide both sides by $-1$ to get $x=-\ln 6\textsf{.}$

Take $\log$ of both sides: $x\log 3=\log 18$. Divide by $\log 3$ to get $x=\dfrac{\log 18}{\log 3}\textsf{.}$

Divide by $2$ to get $e^{3x}=5$. Take $\ln$: $3x=\ln 5$. Divide by $3$ to get $x=\dfrac{\ln 5}{3}\textsf{.}$

Take $\ln$ of both sides: $\dfrac{x}{2}=\ln 9$. Multiply both sides by $2$ to get $x=2\ln 9\textsf{.}$

Divide both sides by $7$ to get $e^{x}=4$. Taking $\ln$ of both sides gives $x=\ln 4\textsf{.}$