Question

Simplify: (1/6)ln abs(y+1)-(1/2)ln abs(y-1)+(1/3)ln abs(y-2).

Answer 1

Please help me.

Answer 2

Is this:

\[\frac{1}{6}\ln (|y+1|) - \frac{1}{2} \ln(|y-1|) + \frac{1}{3}\ln (|y-2|)\]

Answer 3

Yes.

Answer 4

Actually I don't know how to reach there but let me provide what I know:

\[a \ln(x) = \ln(x^a)\]
\[\ln(x) + \ln(y) = \ln(xy)\]
\[\ln(x) - \ln(y) = \ln(\frac{x}{y})\]

Answer 5

You should try to use these formulae there..

Answer 6

So I got (y+1)^(1/6)*(y-1)^(-1/2)*(y-2)^(1/3), is this right?

Answer 7

But this seems to be going nowhere.. :( Sorry..

Answer 8

How can I simplify this more?

Answer 9

@myininaya

Answer 10

\[\frac{1}{6}\ln|y+1| - \frac{1}{2} \ln|y-1| + \frac{1}{3}\ln |y-2|\]
Factor out \(\dfrac{1}{6}\) and apply the properties:
\[\frac{1}{6}\left(\ln|y+1| - \frac{6}{2} \ln|y-1| + \frac{6}{3}\ln |y-2|\right)\\
\frac{1}{6}\left(\ln|y+1| - \ln|y-1|^3 + \ln (y-2)^2\right)\\
\frac{1}{6}\ln\left|\frac{(y+1)(y-2)^2}{(y-1)^3}\right|\]