How do you solve 4e^(x - 1) = 64?

We have that

$4 \cdot {e}^{x - 1} = 64$

$\frac{4 \cdot {e}^{x - 1}}{4} = \frac{64}{4}$ (Divide by 4 both sides)

${e}^{x - 1} = 16$

$\left(x - 1\right) \cdot \ln e = \ln 16$ (Take logarithms in both sides)

$\left(x - 1\right) = 4 \cdot \ln 2$ (Remember that $16 = {2}^{4}$)

$x = 1 + 4 \cdot \ln 2$