College Algebra (11th Edition)

$x=\left\{ -\dfrac{8}{3},0 \right\}$
Changing to exponential form, the given expression, $\log_5[(3x+5)(x+1)]=1 ,$ is equivalent to \begin{array}{l}\require{cancel} (3x+5)(x+1)=5^1 .\end{array} Using concepts of quadratic equations, the solution/s to the equation above is/are \begin{array}{l}\require{cancel} 3x(x)+3x(1)+5(x)+5(1)=5 \\\\ 3x^2+3x+5x+5-5=0 \\\\ 3x^2+8x=0 \\\\ x(3x+8)=0 .\end{array} Equating each factor to zero and then solving for the variable, the solutions are $x=\left\{ -\dfrac{8}{3},0 \right\} .$ Upon checking, $x=\left\{ -\dfrac{8}{3},0 \right\}$ satisfy the original equation.