Helping with long inverse equation

**el123** · August 8th 2009, 05:58 PM

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?

**skeeter** · August 8th 2009, 06:30 PM

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?

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

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

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

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

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

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

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

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

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

**el123** · August 8th 2009, 07:43 PM

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

**11rdc11** · August 8th 2009, 08:06 PM

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

**skeeter** · August 9th 2009, 05:31 AM

Originally Posted by el123

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

my mistake ... I'll edit it.

**el123** · August 9th 2009, 11:00 AM

Thanks again skeeter!

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