College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 4, Exponential and Logarithmic Functions - Section 4.5 - Exponential and Logarithmic Functions - 4.5 Exercises - Page 404: 5


The solution is:$x=\left\{\dfrac{3}{2}\right\}$.

Work Step by Step

Since $1-5^0$, the given equation is equivalent to: $5^{2x-3}=5^0$ Use the property $a^m = a^n \longrightarrow m=n$ to obtain: $\begin{array}{ccc} &2x-3&=&0 \\&2x-3+3&=&0+3 \\&2x&=&3 \\&\dfrac{2x}{2} &= &\dfrac{3}{2} \\&x&=&\dfrac{3}{2} \end{array}$ Thus, the solution is:$x=\left\{\dfrac{3}{2}\right\}$.
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