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#.