1. ## Solve this Logarithm

3ln(x) - 2ln(y) = 4

Express y in terms of x

i think that next it would become

ln (x)^3 - ln(y)^2 = 4

But I really am not sure what to do, so if someone could help me out please, that would be great

2. Start by getting all your $\displaystyle x$ terms on one side and $\displaystyle y$ terms on the other.

$\displaystyle 2\ln{y} = 3\ln{x} - 4$.

Can you go from here?

3. Originally Posted by Prove It
Start by getting all your $\displaystyle x$ terms on one side and $\displaystyle y$ terms on the other.

$\displaystyle 2\ln{y} = 3\ln{x} - 4$.

Can you go from here?
ln(y)^2 = ln(x)^3 -4

in know that you can eliminate ln, however I'm not sure what to do with the subtraction of 4 on the end

4. Originally Posted by gbooker
ln(y)^2 = ln(x)^3 -4

in know that you can eliminate ln, however I'm not sure what to do with the subtraction of 4 on the end
You don't need to use this logarithm rule...

Just remember that the opposite of multiplying by 2 is dividing by 2, and the opposite of a logarithm is an exponential.

5. Originally Posted by Prove It
You don't need to use this logarithm rule...

Just remember that the opposite of multiplying by 2 is dividing by 2, and the opposite of a logarithm is an exponential.
I must be having a really bad mental blank or something I just can't figure it out

6. Originally Posted by gbooker
3ln(x) - 2ln(y) = 4, Express y in terms of x
You can write it as $\ln\left(\frac{x^3}{y^2}\right)=4.$

Recall that $\ln(a)=b$ can be written as $a=e^b$.