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March 29, 2015

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Posted by **marcus** on Wednesday, November 21, 2012 at 3:18am.

g(x)= -e^(x+1) +1, how do I turn this equation into exponential form?

Or can I? I always get confused on what to do especially when I need to the x part of the equation out of the exponent.

- precalculus -
**drwls**, Wednesday, November 21, 2012 at 5:34amg-1 = -e^(x+1)

1-g = e^(x+1)

x + 1 = ln (1 - g)

x = ln(1-g) -1

g will always be less than 1.

Is that what you want? I don't know what you mean by "exponential form"

You can always sketch the graph using the original equation. Just assume different x values, compute g for each, and plot the result

- precalculus -
**marcus**, Wednesday, November 21, 2012 at 6:47amMy book says a way for me to graph/plot points for my function is to write it in exponential form. For example. "Graph (fx)= log base 2(x-2)

y=log base 2 (x-2)

x-2=2^y

x=2^y + 2

So now I can just plug any y value in to get my x value to graph my translated graph. So I was wondering on how to be able to put the above function in exponential form as well? I know when you're done with the graph it's supposed to go through the point (-1,0) and the asymptote is y=1, but I don't know how to get there.

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