## Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole

# Chapter 6 - Logarithmic Functions - 6.6 Solving Logarithmic Equations - 6.6 Exercises: 67

#### Answer

$t = 0, -1, -5$

#### Work Step by Step

$t^{3} +6t^{2} = - 5t$ $t^{3} +6t^{2} + 5t = 0$ $t[t^{2} + 6t + 5] = 0$ $t[t^{2} + 5t + t + 5] = 0$ $t[t(t+5)+1(t+5)] = 0$ $t(t+1)(t+5) = 0$ $t = 0, -1, -5$ Check: When $t = 0$ $t^{3} +6t^{2} \overset{?}{=} - 5t$ $0^{3} +6(0)^{2} \overset{?}{=} - 5(0)$ $0 +0 \overset{?}{=} (0)$ $0 = 0$ When $t = -1$ $t^{3} +6t^{2} \overset{?}{=} - 5t$ $(-1)^{3} +6(-1)^{2} \overset{?}{=} - 5(-1)$ $-1 +6(1) \overset{?}{=} 5$ $-1 +6 \overset{?}{=} 5$ $5 = 5$ When $t = -5$ $t^{3} +6t^{2} \overset{?}{=} - 5t$ $(-5)^{3} +6(-5)^{2} \overset{?}{=} - 5(-5)$ $-125 +6(25) \overset{?}{=} 25$ $-125 +150 \overset{?}{=} 25$ $25 = 25$

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