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 474: 57

Answer

$x = -\frac{1}{2}$

Work Step by Step

Rewrite $\frac{1}{5}$ as an exponential expression: $25^{x} = \frac{1}{5^{1}}$ Since the exponent is in the denominator, rewrite the exponential expression using a negative exponent to get rid of the fraction (use the rule $\frac{1}{a^m}=a^{-m}$): $25^{x} = 5^{-1}$ Rewrite such that both terms have the same base (note that $25=5^2$): $(5^{2})^{x} = 5^{-1}$ When raising a power to a power, multiply the exponents, keeping the base as-is (use the rule $\left(a^m\right)^n=a^{mn}$: $5^{2x} = 5^{-1}$ If two numbers having the same base are equal, that means that their exponents are also the same, so set the exponents equal to one another to solve for $x$: $2x = -1$ Divide both sides by $2$ to solve for $x$: $x = -\frac{1}{2}$
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