Chapter 3 - Section 3.4 - Exponential and Logarithmic Equations - Exercise Set - Page 489: 78

Solution set = $\{5\}.$

Work Step by Step

For the equation to be defined, logarithms must have positive arguments $\left[\begin{array}{llll} 5x+1 \gt 0, & and & 2x+3 \gt 0 & \\ x \gt -1/5 & & x\gt-3/2 & \end{array}\right]$ $x \gt -1/5 \qquad (*)$ This is the condition eventual solutions must satisfy. $\log(5x+1)=\log(2x+3)+\log 2\qquad$... RHS: product rule $\log(5x+1)=\log[(4x+6)\cdot 2]\quad$ ... $\log$ is one-to-one $5x+1=4x+6$ $5x-4x=6-1$ $x=5\qquad$ ... satisfies (*) Thus, the solution set = $\{5\}.$

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