1. ## Help with logarithms

I am really stumped on this question. I have tried re arranging it but I can't seem to get it right.

ln x - ln x^2 = ln 27

2. ## Re: Help with logarithms

logarithms
ln a - ln b= ln a/b
ln 1= 0
ln a +ln b= ln ab
ln e= 1
ln a= ln b is a=b

3. ## Re: Help with logarithms

logarithms
ln a - ln b= ln a/b
ln 1= 0
ln a +ln b= ln ab
ln e= 1
ln a= ln b is a=b

4. ## Re: Help with logarithms

sorry for the double on posts.

5. ## Re: Help with logarithms

Not much help. I know the rules. I just can't solve it. Running out of time.

6. ## Re: Help with logarithms

Originally Posted by SplashDamage
Not much help. I know the rules. I just can't solve it. Running out of time.
At what point do you get stuck? Are you able to combine the left-hand side into a single logarithm? After that you can equate the expressions inside the logs (or, equivalently, make both sides exponential to the base $e$ so that the logarithm operations get canceled out).

7. ## Re: Help with logarithms

I'm interested in knowing how to solve this one too.

8. ## Re: Help with logarithms

Originally Posted by SplashDamage
I am really stumped on this question. I have tried re arranging it but I can't seem to get it right.

ln x - ln x^2 = ln 27
With the rule $\ln(a)-\ln(b)=\ln\left(\frac{a}{b}\right)$ the equation is equivalent to:
$\ln\left(\frac{x}{x^2}\right)=\ln(27)$
$\Leftrightarrow \ln\left(\frac{1}{x}\right)=\ln(27)$
Therefore $\frac{1}{x}=27$ and thus $x=\frac{1}{27}$ is the solution.