Precalculus (6th Edition) Blitzer

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

Chapter 10 - Cumulative Review Exercises - Page 1127: 12


The value of ${{\log }_{2}}x+{{\log }_{2}}\left( 2x-3 \right)=1$ is $ x=2$.

Work Step by Step

We have to find the value of x from ${{\log }_{2}}x+{{\log }_{2}}\left( 2x-3 \right)=1$ by evaluating the left-hand side as follows: $\begin{align} & {{\log }_{2}}x+{{\log }_{2}}\left( 2x-3 \right)=1 \\ & {{\log }_{2}}x\left( 2x-3 \right)=1 \\ & {{\log }_{2}}\left[ 2{{x}^{2}}-3x \right]=1 \end{align}$ Then, using if ${{p}^{q}}=r $, then ${{\log }_{p}}r=q $, one gets: $\begin{align} & 2{{x}^{2}}-3x={{2}^{1}} \\ & 2{{x}^{2}}-3x-2=0 \\ & 2{{x}^{2}}-4x+x-2=0 \end{align}$ Simplify it further: $\begin{align} & 2x\left( x-2 \right)+x-2=0 \\ & \left( 2x+1 \right)\left( x-2 \right)=0 \end{align}$ So, $ x=2$ or $ x=-\frac{1}{2}$. If $ x=-\frac{1}{2}$, then ${{\log }_{2}}x $ does not exist as the logarithm function does not take on negative values. Therefore, the only possible value of x is 2. Thus, the value is $ x=2$.
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