Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 7 - Exponential and Logarithmic Functions - 7-5 Exponential and Logarithmic Equations - Practice and Problem-Solving Exercises - Page 473: 13



Work Step by Step

Note that $4=2^2$ so the given equation is equivalent to: $2^{3x}=4^{x+1}$ $2^{3x}=(2^2)^{x+1}$ $2^{3x}=2^{2(x+1)}$ $2^{3x}=2^{2x+2}$ Using the rule $a^m=a^n\longrightarrow m=n$ gives: $3x=2x+2$ $3x-2x=2$ $x=2$ Check: $2^{3*2}=4^{2+1}$ $2^{6}=4^{3}$ $64=64$ Thus, the answer is correct.
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