College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Section 4.3 - Logarithmic Functions - Summary Exercises on Inverse, Exponential, and Logarithmic Functions - Page 427: 44


$x=\text{ any real number}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 5^{2x-6}=25^{x-3} ,$ use the laws of exponents to express both sides in the same base. Then equate the exponents. $\bf{\text{Solution Details:}}$ Using exponents, the given equation is equivalent to \begin{array}{l}\require{cancel} 5^{2x-6}=(5^2)^{x-3} .\end{array} Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the equation above is equivalent to \begin{array}{l}\require{cancel} 5^{2x-6}=5^{2(x-3)} .\end{array} Since the bases are the same, then the exponents can be equated. That is, \begin{array}{l}\require{cancel} 2x-6=2(x-3) \\\\ 2x-6=2(x)+2(-3) \\\\ 2x-6=2x-6 \\\\ 2x-2x=-6+6 \\\\ 0=0 \text{ (TRUE)} .\end{array} Since the equation above ended with a TRUE statement, the given equation is an identity. Hence, the values of $x$ that satisfy the given equation is $ x=\text{ any real number} .$
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