College Algebra, Chapter 5, 5.4, Section 5.4, Problem 48

Solve the Logarithmic Equation $\log_5 x + \log_5 ( x + 1) = \log_ 5 20$ for $x$.

$
\begin{equation}
\begin{aligned}
\log_5 x + \log_5 ( x + 1) &= \log_ 5 20\\
\\
\log_5 x (x + 1) &= \log_5 20 && \text{Laws of Logarithms } \log_a AB = \log_a A + \log_a B\\
\\
5^{\log_5 x ( x + 1)} &= 5^{\log_5 20} && \text{Raise 5 to each side}\\
\\
x (x + 1) &= 20 && \text{Property of } \log\\
\\
x^2 + x &= 20 && \text{Distributive Property}\\
\\
x^2 + x - 20 &= 0 && \text{Subtract 20}\\
\\
(x - 4)(x + 5) &= 0
\end{aligned}
\end{equation}
$

Solve for $x$

$
\begin{equation}
\begin{aligned}
x -4 &= 0 &&\text{and}& x + 5 &= 0 \\
\\
x &= 4 &&& x &= -5
\end{aligned}
\end{equation}
$

The only solution in the given equation is $x = 4$, since $x = -5$ will make the term a negative value.