how do I solve: ln(x)(3t+1)=ln(y)(2t+1) for y?

Question

campbell_st · Answer

use change of base either base e or base 10

the change for base formula is 



$$\log_{x}(b) = \frac{\ln(b)}{\ln(x)}$$

once you use change on base all terms are the same base so you then start to solve for y

phi · Answer

first step, divide both sides by (2t+1)

phi · Answer

then make each side the exponent of e

john_es · Answer

$$\ln(x^{3t+1})=\ln(y^{2t+1})\Rightarrow x^{3t+1}=y^{2t+1}\Rightarrow y=x^{\frac{3t+1}{2t+1}}$$

anonymous · Answer

i have $$\ln(x)*(1+3t) = \ln(y)*(1+2t)$$
so you say I should do make that
$$\frac{ \ln(x)(1+3t) }{ (1+3t)}=\ln(y)$$

anonymous · Answer

sorry (1+2t) on the bottom

phi · Answer

now make each side the exponent of e
use the rule
$$ e^{\ln(y)}= y $$

on the left side  you get
$$ e^{\ln(x) A} $$
where A is (1+3t)/(1+2t)

the left side can be written as
$$ \left(e^{\ln(x)} \right)^A $$
now use the same rule as you did for the right side

phi · Answer

you should get the answer John_ES has already posted up above.