Solve this Logarithm

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March 13th 2011, 10:29 PM
gbooker

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
March 13th 2011, 10:34 PM
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?
March 13th 2011, 10:43 PM
gbooker

Quote:

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
March 13th 2011, 10:45 PM
Prove It

Quote:

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.
March 13th 2011, 11:30 PM
gbooker

Quote:

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
March 14th 2011, 03:34 AM
Plato

Quote:

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 $a=e^b$ .

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