How do you solve $log(x+2)+log(x-2)=1$ ?

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Answer 1 · 2016-02-25T22:49:51

$x=sqrt14$

Simplify the left hand side through the logarithm rule:

$log(a)+log(b)=log(ab)$

Thus, we obtain

$log[(x+2)(x-2)]=1$

Distributed, this gives

$log(x^2-4)=1$

Now, recall that $log(a)$ really means $log_10(a)$ :

$log_10(x^2-4)=1$

To undo the logarithm, exponentiate both sides with base $10:$

$10^(log_10(x^2-4))=10^1$

$x^2-4=10$

Solve:

$x^2=14$

$x=+-sqrt14$

Be very careful when solving logarithm functions--always plug your answer back into the original expression.

Note that the solution $x=-sqrt14$ is invalid, since it would make $log(x-2)$ have a negative argument, and $log(a)$ only exists from $a>0$ .

Thus, the only valid solution is $x=sqrt14$ .

How do you solve log(x+2)+log(x-2)=1 log(x+2)+log(x-2)=1?