# How do you solve log_ 5 (x+4) + log _ 5 (x+1)= 2?

Feb 17, 2016

You can start by simplifying using the log property ${\log}_{a} n + {\log}_{a} m = {\log}_{a} \left(n \times m\right)$

#### Explanation:

${\log}_{5} \left(x + 4\right) + {\log}_{5} \left(x + 1\right) = 2$

${\log}_{5} \left(x + 4\right) \left(x + 1\right) = 2$

${\log}_{5} \left({x}^{2} + 4 x + x + 4\right) = 2$

Convert to exponential form:

${x}^{2} + 5 x + 4 = 25$

${x}^{2} + 4 x - 21 = 0$

We can solve this equation by factoring. This is a trinomial of the form $y = a {x}^{2} + b x + c$. To factor this, you must find two numbers that multiply to c and that add to b. Two numbers that multiply to -21 and that add to 4 are +7 and -3.

$\left(x + 7\right) \left(x - 3\right) = 0$

$x = - 7 \mathmr{and} 3$

Since having a negative log is undefined, the only solution to the equation is x = 3.

Hopefully this helps.