Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 7 Exponential and Logarithmic Functions - 7.6 Solve Exponential and Logarithmic Equations - 7.6 Exercises - Skill Practice - Page 519: 4

Answer

x = 2

Work Step by Step

Since $49 = 7^2$ we can multiply the exponent of the 49 hand side by $2$ while reducing the base to $7$. The new equation is $7^{3x+4}= 7^{2(2x+1)}$. Distributing the $2$ gives, $7^{3x+4}= 7^{4x+2}$ We can now use Key Concept Algebra on pg. 515 stating If b is a positive number other than 1, then $b^x $ $= b^y$ if and only if $x = y$. Thus $ 3x+4 = 4x+2$ After moving the variables to one side and the constants to the other, $ -2 = -x $ or $x=2$
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