Question

how would i solve log1/3(x^2+x)-log1/3(x^2-x) =-1

Answer 1

\[\LARGE{\log \frac{1}{3}(x^2+x)-\log \frac{1}{3}(x^2-x)=-1}\]Is this ur question ???
@Modelcouture

Answer 2

\[\LARGE{\log \frac{{1 \over 3}(x^2+x)}{{1 \over 3}(x^2-x)}=-1}\]\[\LARGE{\color{red}{\text{Since} \space \log_a b-\log_ad=\log_a \frac{b}{d}}} \]

Answer 3

@hartnn am i right ??

Answer 4

I'm nt totally sure :/

Answer 5

what you did is correct, but is that the question ? i doubt...

Answer 6

He need this equation to solve :)

Answer 7

i think base is (1/3)

Answer 8

This is the problem that we couldn't understand the question point to point :(

Answer 9

\[\LARGE{Log_\frac{1}{3}(x^2+x)-Log_\frac{1}{3}(x^2-x)=-1}\]\[\LARGE{Log_\frac{1}{3}\frac{x+1}{x-1}=-1}\]\[\LARGE{\frac{1}{3}^{-1}=\frac{x+1}{x-1}}\]\[\LARGE{3=\frac{x+1}{x-1}}\]Solve for x now :)

Answer 10

You gt the question correctly @hartnn :)