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

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Answer 1 · 2016-08-11T02:29:53

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

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).

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