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Suppose i had this equation , which i do , well i have one similar.
I keep getting stuck on finding the inverse becuase of so many things in it , i am not sure about how where to take the ln of the euqaiotn to get rid of the e.
$\displaystyle f(x)= 40+30(2-e^{(-0.2x)}) $
Could some one walk me through it?
$\displaystyle y = 40+30(2 - e^{-0.2x})$Quote:
Originally Posted by el123
$\displaystyle y = 40 + 60 - 30e^{-0.2x}$
$\displaystyle y = 100 - 30e^{-0.2x}$
$\displaystyle x = 100 - 30e^{-0.2y}$
$\displaystyle 100-x = 30e^{-0.2y}$
$\displaystyle \frac{100-x}{30} = e^{-0.2y}$
$\displaystyle \ln\left(\frac{100-x}{30}\right) = -0.2y$
$\displaystyle -5\ln\left(\frac{100-x}{30}\right) = y$
$\displaystyle y = \ln\left(\frac{30}{100-x}\right)^5$
Why did you switch the sign around when you exchanged y and x?
Thats what you have to do to find an inverse. Switch y and x. To check to see if you found the inverse plug in your new equation into the original and it should equal y=x
my mistake ... I'll edit it.Quote:
Originally Posted by el123
Thanks again skeeter!
