Chapter 4 - Exponential and Logarithmic Functions - Section 4.4 Logarithmic Functions - 4.4 Assess Your Understanding - Page 322: 102

$\dfrac{1-\ln{13}}{2}$

$\because e^y=x \text{ is equivalent to } y=\ln{x}$ $\therefore e^{-2x+1} = 13 \text{ is equivalent to } -2x+1=\ln{13}$ Solve the equation above to obtain: \begin{align*}-2x+1&=\ln{13}\\\\ -2x&=\ln{13}-1\\\\ \frac{-2x}{-2}&=\frac{\ln{13}-1}{-2}\\\\ x&=-\frac{\ln{13}}{2}+\frac{1}{2}\\\\ x&=\frac{1}{2}-\frac{\ln{13}}{2}\\\\ x&=\dfrac{1-\ln{13}}{2}\end{align*}