# How do you solve log_3 x + log_3 (x + 4) = log_3 32 ?

May 15, 2016

we use the additive property of logarithms to simplify the equation to a quadratic and find the solution $x = 4$.

#### Explanation:

We really want to take the anti-log of both sides of this equation, but one side has a sum of two logarithms. To solve this we need to use the property of logarithms that says that adding two logarithms is the same as the logarithm of the two arguments multiplied together:

$\log \left(a \cdot b\right) = \log \left(a\right) + \log \left(b\right)$

So our equation can be re-written as

${\log}_{3} \left(x \cdot \left(x + 4\right)\right) = {\log}_{3} 32$

Now we can take the anti-log, which is simply raising both sides of the equation as the exponent of 3:

${3}^{{\log}_{3} \left(x \cdot \left(x + 4\right)\right)} = {3}^{{\log}_{3} 32}$

which simplifies to

$x \cdot \left(x + 4\right) = 32$

which gives

${x}^{2} + 4 x - 32 = 0$

Now we can factor to give

$\left(x + 8\right) \left(x - 4\right) = 0$

$x = 4 , - 8$
However, one of these solutions is spurious, since if $x = - 8$ then we would need to take the logarithm of a negative number in our original formula. Therefore, the only actual solution is
$x = 4$