# Single Variable Calculus, Chapter 7, 7.3-2, Section 7.3-2, Problem 14

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Determine the solution of the equation a.) $\ln (1 + \sqrt{x}) = 2$ and b.) $e^{\frac{1}{(x - 4)}} = 7$ correct to four decimal places.

a.) $\ln ( 1 + \sqrt{x}) = 2$

\begin{aligned} & e^{\ln (1 + \sqrt{x})} = e^2 \\ \\ & 1 + \sqrt{x} = e^2 \\ \\ & \sqrt{x} = e^2 - 1 \\ \\ & x = (e^2 - 1)^2 \\ \\ & x = 40.8200 \end{aligned}

b.) e^{\frac{1}{(x - 4)} } = 7

\begin{aligned} & \lne^{\frac{1}{(x - 4)}} = \ln 7 \\ \\ & \frac{1}{x - 4} = \ln 7 \\ \\ & \frac{1}{\ln 7} = x - 4 \\ \\ & x = \frac{1}{\ln 7} + 4 \\ \\ & x = 4.5139 \end{aligned}