## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 3 - Section 3.4 - Exponential and Logarithmic Equations - Exercise Set: 29

#### Answer

$x \approx 1.53$

#### Work Step by Step

Divide both sides of the equation by 5to obtain $e^x=\dfrac{23}{5}.$ The base in the exponential equation is $e$ so take the natural logarithm on both sides to obtain $\ln{e^x}=\ln{(\frac{23}{5})}.$ Use the inverse property $\ln{e^x}=x$ on the left side to obtain $x = \ln{(\frac{23}{5})}.$ Use a calculator and round-off the answer to two decimal places to obtain $x \approx 1.53.$

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