## Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole

# Chapter 6 - Logarithmic Functions - 6.6 Solving Logarithmic Equations - 6.6 Exercises: 44

#### Answer

$x = 3$

#### Work Step by Step

$\log_6 (4x) + \log_6 x = 2$ $\log_6 (4x^{2}) = 2$ $6^{2} = 4x^{2}$ $36 = 4x^{2}$ $9 = x^{2}$ $x = ±\sqrt 9$ $x = ±3$ Since $x$ can't be negative, the only possible solution is $x = 3$. This is because $\log$ of a negative number does not exist and therefore we only accept the positive number of $x$. Check: $\log_6 (4(3)) + \log_6 3 \overset{?}{=} 2$ $\log_6 (12) + \log_6 3 \overset{?}{=} 2$ $\log_6 (12 \times 3) \overset{?}{=} 2$ $\log_6 (36) \overset{?}{=} 2$ $\log_6 (6^{2}) \overset{?}{=} 2$ $2\log_6 (6) \overset{?}{=} 2$ $2(1) \overset{?}{=} 2$ $2 = 2$

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