Helping with long inverse equation

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Aug 8th 2009, 05:58 PM
el123

Helping with long inverse equation

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.

$f(x)= 40+30(2-e^{(-0.2x)})$

Could some one walk me through it?
Aug 8th 2009, 06:30 PM
skeeter

Quote:

Originally Posted by el123

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.

$f(x)= 40+30(2-e^{(-0.2x)})$

Could some one walk me through it?

$y = 40+30(2 - e^{-0.2x})$

$y = 40 + 60 - 30e^{-0.2x}$

$y = 100 - 30e^{-0.2x}$

$x = 100 - 30e^{-0.2y}$

$100-x = 30e^{-0.2y}$

$\frac{100-x}{30} = e^{-0.2y}$

$\ln\left(\frac{100-x}{30}\right) = -0.2y$

$-5\ln\left(\frac{100-x}{30}\right) = y$

$y = \ln\left(\frac{30}{100-x}\right)^5$
Aug 8th 2009, 07:43 PM
el123

Why did you switch the sign around when you exchanged y and x?
Aug 8th 2009, 08:06 PM
11rdc11

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
Aug 9th 2009, 05:31 AM
skeeter

Quote:

Originally Posted by el123

Why did you switch the sign around when you exchanged y and x?

my mistake ... I'll edit it.
Aug 9th 2009, 11:00 AM
el123

Thanks again skeeter!