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

Published by Pearson

# Chapter 4 - Exponential and Logarithmic Functions - Section 4.6 Logarithmic and Exponential Equations - 4.6 Assess Your Understanding - Page 337: 65

#### Answer

No real solutions.

#### Work Step by Step

Re-write first term as: $4^x=(2^{2})^{x}=(2^{x})^{2}$ We wish to substitute $a=2^{x}$, so the equation becomes: $3a^{2}+4a+8=0$ This gives a quadratic equation, whose factors can be found by using the quadratic formula. $a=\dfrac{−4 \pm \sqrt {4^2-(4)(3)(8)}}{(2)(3)} = 42−4(3)(8)=−80$ We see the discriminant is negative, so there are no real solutions.

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