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: 3


$x = 8$

Work Step by Step

Since $25 = 5^2$ we can multiply the exponent of the left hand side by $2$ while reducing the base to $5$. The new equation is $5^{x-4}= 5^{2(x-6)}$. Distributing the $2$ gives, $5^{x-4}= 5^{2x-12}$ 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 $ x -4 = 2x-12$ After moving the variables to one side and the constants to the other, $ 8 = x $ or $x=8$
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