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#.