Question

Write
2[ln x − ln(x + 1) − ln(x − 1)]
as the logarithm of a simple quantity

Answer 1

\[2\ln [ x-[\ln (x+1)+\ln (x-1)]]=2\ln [\frac{ x}{(x+1)(x-1)}]=\ln \frac{x}{(x+1)(x-1)}]^2\]

Answer 2

Last part got cut off.
\[\ln [\frac{x}{(x+1)(x-1)}]^{2}\]

Answer 3

A few neat properties of logarithms to know about...
ln(a) - ln(b) = ln( a/b )
a * ln(b) = ln(b^a)

2 [ ln(x) - ln(x + 1) - ln(x - 1) ]
Let's work with what is inside of those square brackets first and we'll replace it later.

ln(x) - ln(x + 1) - ln(x - 1) I'll rewrite with first two terms in parentheses (its not needed, I do it to show process.)
= ( ln(x) - ln(x + 1) ) - ln(x - 1) Here, we can use that first property.
= ln(x / (x+1)) - ln(x - 1) And then again, use that property.
= ln( (x / (x + 1)) / (x - 1) ) a/b = a*(1/b)
= ln( (x / (x + 1)) * (1 / (x - 1)) ) We can see that (x + 1)(x -1) will become x^2 - 1
= ln( x / (x^2 - 1) )

Now, we'll replace that stuff in the brackets with what we found.

2[ ln( x / (x^2 - 1) ) ] Here's where the second property listed comes in.
ln( (x / (x^2 - 1))^2 )