Question

Express
Log-2logx+3log (x+1) -log (x^2 -1)
as a single logarithm

Answer 1

With a few properties of logarithms you should be able to combine that into a single logarithm:

\[\log a+\log b = \log a*b\]\[\log a - \log b = \log \frac{a}{b}\]\[\log x^a = a\log x\]

Answer 2

thanks whpalmer4

Answer 3

\(\large\color{blue}{ \sf Log(-2)~log(x)+3log (x+1) -log (x^2 -1)}\)

Answer 4

(hate disconnections)

Answer 5

Is that supposed to be \[-2\log x + 3\log(x+1) - \log(x^2-1)\]?
I'm confused by the initial "Log-"...you can't take the logarithm of a negative number if you expect a real answer.

Answer 6

( without missing the 1st 3 letters of the questions in the blue box )

\(\large\color{blue}{ \sf Log(-2)~log(x)+3log (x+1) -log (x^2 -1)}\)

Answer 7

it can't be just log, so I guess the question is what I wrote, starting from log(-2)

Answer 8

Tell me, what power do you raise 10 to to get a negative number? :-)

Answer 9

\[\log_b x =a\]means that \[b^a=x\]\[10^a=-2\]\[a=\]

Answer 10

https://www.google.com/#q=log(-2) WOW holy cow, that's weird!

Answer 11

idk

Answer 12

@Imtiaz7 can you re-write the question as clearly as possible please ?

Answer 13

The natural log is more, well, natural:

\[\ln -2 = i\pi + \ln 2 \approx 0.693147+3.14159i\]

Your manipulation wasn't quite correct.
\[\log_{10}x = -2\rightarrow 10^{-2} = x = \frac{1}{10^2} = 0.01\]

But that's not the problem I posed to you. \[\log_{10} -2 = x\rightarrow 10^x = -2\]