Suppose we want to find the solutions to this equation:

${\log }_{16}\left(x\+13\right)\+{\log }_{16}\left(x-2\right)\=1$

Use the (reverse of the) product rule for logarithms:

${\log }_{16}\left(\left(x\+13\right)\cdot \left(x-2\right)\right)\=1$

Expand the inner quadratic:

${\log }_{16}\left(x^2\+11\mathbf{}x-26\right)\=1$

Each side of the above equation is graphed to the right. The solution to the problem is their intersection. Note that since we can only take the logarithm of positive numbers, the domain of the left-hand side function is the set of all numbers satisfying both $x\+13\>0$ and $x-2\>0$, or equivalently $x\>-13$ and $x\>2$. This means the domain of the function $y\={\log }_{16}\left(x\+13\right)\+{\log }_{16}\left(x-2\right)$ is the set of all numbers satisfying $x\>2$.

Inverting the logarithm by raising 16 to both sides gives:

$x^2+11 x-26\={16}^1$

Rewrite this as:

$x^2+11 x-42\=0$

This factors as:

$\left(x\+14\right)\cdot \left(x-3\right)\=0$, so$x\=-14$ or $x\=3$

Graphed on the right is the function $x^2\+11 x-42$, with the two roots shown. However, due to the restrictions placed on the original equation, that is, $x\>2$, only solutions found in the solid region are valid for the original problem. So the solution to the original problem is $x\=3$.