# log_3(x+12)+log_3(x-12)=4 ?

$x = 15$

#### Explanation:

We have:

${\log}_{3} \left(x + 12\right) + {\log}_{3} \left(x - 12\right) = 4$

We can combine the left side by saying:

${\log}_{3} \left(\left(x + 12\right) \left(x - 12\right)\right) = 4$

Know that we can rewrite log questions using the following format:

${a}^{b} = c \iff {\log}_{a} c = b$

So we can write out question as:

${3}^{4} = \left(x + 12\right) \left(x - 12\right)$

And let's expand:

$81 = {x}^{2} - 144$

${x}^{2} = 225$

$x = \pm 15$

And now let's test the solutions:

For $x = 15$

${\log}_{3} \left(x + 12\right) + {\log}_{3} \left(x - 12\right) = 4$

${\log}_{3} \left(15 + 12\right) + {\log}_{3} \left(15 - 12\right) = 4$

${\log}_{3} \left(27\right) + {\log}_{3} \left(3\right) = 4$

$3 + 1 = 4$

For $x = - 15$

${\log}_{3} \left(- 15 + 12\right) + {\log}_{3} \left(- 15 - 12\right) = 4$

${\log}_{3} \left(- 3\right) + {\log}_{3} \left(- 27\right) = 4$ and this doesn't work because we can't have negatives inside the log function.

Therefore our solution is $x = 15$

Oct 11, 2016

$x = 15$

#### Explanation:

I have assumed that you meant ${\log}_{3}$ as I have now shown in the question.

First problem:
You can't mix log terms and number terms, all one or all the other.

Second problem:
How do you write 4 as ${\log}_{3}$?

Compare:
${10}^{1} = 10 \Leftrightarrow {\log}_{10} 10 = 1 \text{ and } {3}^{1} = 3 \Leftrightarrow {\log}_{3} 3 =$
${10}^{2} = 100 \Leftrightarrow {\log}_{100} = 2 \text{ and } {3}^{2} = 9 \Leftrightarrow {\log}_{3} 9 = 2$
${10}^{3} = 1000 \Leftrightarrow {\log}_{1000} = 3 \text{ and } {3}^{3} = 27 \Leftrightarrow {\log}_{3} 27 = 3$

${\log}_{3} \left(x + 12\right) + {\log}_{3} \left(x - 12\right) = \textcolor{red}{4} \text{ } \leftarrow$ can be written as:
${\log}_{3} \left(x + 12\right) + {\log}_{3} \left(x - 12\right) = \textcolor{red}{{\log}_{3} 81}$

Condense the two log terms into one:

${\log}_{3} \left(x + 12\right) \left(x - 12\right) = {\log}_{3} 81$

Now use the log law: If $\log A = \log B \Leftrightarrow A = B$

$\cancel{{\log}_{3}} \left(x + 12\right) \left(x - 12\right) = \cancel{{\log}_{3}} 81$

$\left(x + 12\right) \left(x - 12\right) = 81 \text{ } \leftarrow$ notice the difference of squares

${x}^{2} - 144 = 81$

${x}^{2} = 225$

$x = \pm \sqrt{225} = \pm 15$

However in this question, $- 15$ is not valid.

$x = 15$

Be sure to notice the base is 3, not 10!