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

1 Answer
Gió
Oct 28, 2015

I found: #x=+sqrt(11)#

Explanation:

You can use the rule of logs that relates the sum of logs of same base and the multiplication of the integrand as:
#log_ax+log_ay=log_a(x*y)#
and get:

#log_a[(x+1)(x-1)]=1# using the definition of log:

#(x+1)(x-1)=a^1# (1)

Now it depends upon the base #a# of your logs....and also if it the same for both the original ones!
If #a=10# then you have:
#x^2-1=10#
#x^2=11#
#x=+-sqrt(11)#
Where I accept the positive only, #x=+sqrt(11)#

If #a# isn't #10# simply insert the right value into (1) and solve.

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