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: 57


$x =\dfrac{\ln{3}}{\ln{2}}\approx 1.585$

Work Step by Step

We let $t=2^{x}$, so the equation becomes: $$t^{2}+t-12=0 $$ Factor the quadratic trinomial by looking for factors of $-12$ whose sum is $1$ (the numerical coefficient of the middle term). to obtain: $$(t+4)(t-3)=0$$ Use the Zero Product Property by equating each factor to $0$ to obtain the equations: $$ t+4 =0$$ and $$t-3=0$$ Solve each equation to obtain: $$t=-4 \text{ and }t=3$$ Since $t=2^{x}=-4$, then we have either $$2^x=-4\quad \text{ or } \quad 2^x=3$$ Note that $2^{x}=-4$ has no real solutions because when $2$ is raised to any real number, the result will never be $-4$. Hence, we will only consider $2^{x}=3$ To find the value of $x$, we apply $\log$ base $2$ to both sides to obtain: $$\log_{2}2^{x}=\log_{2}3$$ Use the properties $\log_a{a^x}=x$ and the change-of-base formula $\log_a{b}=\frac{\ln{b}}{\ln{a}}$ to solve for $x$: $$x=\log_{2} (3)\\ x=\dfrac{\ln{3}}{\ln{2}}\\ x \approx 1.585$$ Thus, the answer is: $x =\dfrac{\ln{3}}{\ln{2}}\approx 1.585$
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