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.

Related questions
See all questions in Logarithmic Models
Impact of this question
966 views around the world
You can reuse this answer
Creative Commons License