Answer
$log_23 \approx1.585$
Work Step by Step
Step 1. Factor the equation to get $(2^x+4)(2^x-3)=0 \longrightarrow 2^x=-4,3$
Step 2. For $2^x=-4$, there is no real solution.
Step 3. For $2^x=3$, we have $x=log_23=\frac{ln3}{ln2}\approx1.585$
Precalculus (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0-32197-907-9
ISBN 13:
978-0-32197-907-0
Chapter 5 - Exponential and Logarithmic Functions - 5.6 Logarithmic and Exponential Equations - 5.6 Assess Your Understanding - Page 312: 57
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