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
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Start by getting all your $\displaystyle \displaystyle x$ terms on one side and $\displaystyle \displaystyle y$ terms on the other. $\displaystyle \displaystyle 2\ln{y} = 3\ln{x} - 4$. Can you go from here?
Originally Posted by Prove It Start by getting all your $\displaystyle \displaystyle x$ terms on one side and $\displaystyle \displaystyle y$ terms on the other. $\displaystyle \displaystyle 2\ln{y} = 3\ln{x} - 4$. 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
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.
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
Originally Posted by gbooker 3ln(x) - 2ln(y) = 4, Express y in terms of x You can write it as $\displaystyle \ln\left(\frac{x^3}{y^2}\right)=4.$ Recall that $\displaystyle \ln(a)=b$ can be written as $\displaystyle a=e^b$.
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