# Math Help - more logs

1. ## more logs

hiya
well ive been given the formula $(y/x)^a=be^-x$ cant make this end bit look right its to the power of -x
and I have to make it a linear formula...was thinking I'm probably going to have to take logs but was wondering if I should take ln e so I can eliminate the e out

OR does anyone know a better way to make this a linear formula

much help appriciated

2. Originally Posted by p3achy
hiya
well ive been given the formula $(y/x)^a=be^-x$ cant make this end bit look right its to the power of -x
and I have to make it a linear formula...was thinking I'm probably going to have to take logs but was wondering if I should take ln e so I can eliminate the e out

OR does anyone know a better way to make this a linear formula

much help appriciated
Take logs!

$\ln{|\frac{y}{x}^a|} = \ln{|be^{-x}|}$

$\ln{|\frac{y^a}{x^2}|} = \ln{|b|}+\ln{|e^{-x}|}$

$\ln{|y^a|}-\ln{|x^2|} = \ln{|b|}+\ln{|e^{-x}|}$

$\ln{|y^a|}-\ln{|x^2|} = \ln{|b|}-x$

$a\ln{|y|}-ln(x^2) = \ln{|b|}-x$

$a\ln{|y|} = \ln{|x^2|} + \ln{|b|}-x$

$a\ln{|y|} = \ln{|\frac{x^2}{b}|}-x$

$\ln{|y|} = \frac{\ln{|\frac{x^2}{b}|}-x}{a}$

$\ln{|y|} = \frac{1}{a}\times \ln{|\frac{x^2}{b}|}-\frac{x}{a}$

$\displaystyle \ln{|y|} = \ln{|(\frac{x^2}{b})^{\frac{1}{a}}|}-\frac{x}{a}$

$\displaystyle y = e^{\ln{|(\frac{x^2}{b})^{\frac{1}{a}}|}-\frac{x}{a}}$

$\displaystyle y = (\frac{x^2}{b})^{\frac{1}{a}}.\frac{1}{e^{\frac{x} {a}}}$

3. thx looks like i was mostly there just messed me logs up at the end