How do you condense #2[ln 8- ln(x^2+1)]#?

1 Answer
Aug 11, 2016

Answer:

I got as far as: #ln(8/(x^2+1))^2#

Explanation:

We can use two properties of logs to write:
#2[ln(8/(x^2+1))]=#
(change a subtraction of logs into a division of arguments)

and again:
#=ln(8/(x^2+1))^2#
(transport a number multiplying the log as exponent of the argment).