How do you solve $ln(x^2-20) = ln5$ ?

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Answer 1 · 2015-11-30T20:31:49

The equation has 2 solutions: $5$ and $-5$ .

First we have to find the possible values for $x$ (the domain).
Since $log$ is only defined for positive arguments we have to solve inequality:

$x^2-20>0$

$x^2>20$

$abs(x)>sqrt(20)$

$x in (-oo;-2sqrt(5)) uu (2sqrt(5);+oo)$

Now we can solve the equation:

Since the base of the logarythm is the same on both sides we can write this equation as:

$x^2-20=5$

$x^2=25$

$abs(x)=5$

$x=5 vv x=-5$

Both $5$ and $-5$ are in the domain, so the equation has 2 solutions.

Answer:
The equation has 2 solutions: $5$ and $-5$ .

How do you solve ln(x^2-20) = ln5ln(x^2-20) = ln5?