Precalculus (6th Edition) Blitzer

$x \approx -1.14$
The base in the exponential equation is $e$, so take the natural logarithm on both sides to obtain $\ln{e^{1-5x}}=\ln{793}.$ Use the property $\ln{e^b}=b$ (where b=1-5x) on the left side to obtain $1-5x = \ln{793}.$ Solve for $x$: $1-5x = \ln{793} \\-5x=\ln{793} - 1 \\x=\dfrac{\ln{793}-1}{-5}.$ Use a calculator and round-off the answer to two decimal places to obtain $x \approx -1.14.$