Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Cumulative Review Exercises - Page 880: 18


The solution of the equation is $\underline{-2+{{\log }_{3}}11}$

Work Step by Step

Consider the given equation: ${{3}^{x+2}}=11$ Take the logarithm of both sides and obtain: $\log {{\left( 3 \right)}^{x+2}}=\log \left( 11 \right)$ Then use the multiplication and division rules of logarithmic functions to solve for x: $\begin{align} & \log {{\left( 3 \right)}^{x+2}}=\log \left( 11 \right) \\ & \left( x+2 \right)\log \left( 3 \right)=\log \left( 11 \right) \\ & \left( x+2 \right)=\log \left( \frac{11}{3} \right) \\ & \left( x+2 \right)={{\log }_{3}}11 \\ & x={{\log }_{3}}11-2 \\ & x=-2+{{\log }_{3}}11 \end{align}$ Thus the solution of the equation is $ x=-2+{{\log }_{3}}11$.
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