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

1 Answer
Aug 11, 2016

I got as far as: $\ln {\left(\frac{8}{{x}^{2} + 1}\right)}^{2}$

Explanation:

We can use two properties of logs to write:
$2 \left[\ln \left(\frac{8}{{x}^{2} + 1}\right)\right] =$
(change a subtraction of logs into a division of arguments)

and again:
$= \ln {\left(\frac{8}{{x}^{2} + 1}\right)}^{2}$
(transport a number multiplying the log as exponent of the argment).