Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 4 - Exponential and Logarithmic Functions - Section 4.6 Logarithmic and Exponential Equations - 4.6 Assess Your Understanding - Page 337: 66


There are no real solutions.

Work Step by Step

Re-write as: $49^x=(7^{2})^{x}=(7^{x})^{2}$ We wish to substitute $a=7^{x}$, so the equation becomes: $2a^{2}+11a+5=0 $ This gives a quadratic equation, whose factors are: $(2a+1)(a+5)=0$ Use the zero factor property to obtain: $ 2u+1 =0 \implies a=-\dfrac{1}{2}$ and $a+5 =0 \implies a=-5$ Further, we will plug $a=7^x$ into the above value of $a$ to get the value of $x$ as follows: $7^x=\dfrac{-1}{2} $ and $7^x =-5$. Since, $7^x \gt 0$, the solutions of $7^x=\dfrac{-1}{2} $ and $7^x =-5$ cannot be accepted. Thus, there are no real solutions.
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