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October 24th, 2013, 08:07 AM   #1
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Another exponential equation....

Ok, so this time it is:

e^2x - 3e^-x = 2

So far I get:

e^2x - 3(1/3e^x) -3 = 2 ------to take away -x
e^x(e^2x) - 3 = 2
e^3x = 5
lne^3x = ln5
3xlne = ln5
3x = ln5
x = ln5/3

.....is that right????
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October 24th, 2013, 08:16 AM   #2
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Re: Another exponential equation....

Quote:
e^2x - 3e^-x = 2
Are you sure it's not e^x - 3e^-x = 2?
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October 24th, 2013, 08:22 AM   #3
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Re: Another exponential equation....

I wrote the problem wrong....should be...

e^2x - 3e^x - 10 = 0

...more than wrong, really wrong...
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October 24th, 2013, 08:28 AM   #4
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Re: Another exponential equation....

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October 24th, 2013, 08:38 AM   #5
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Re: Another exponential equation....

gregs method is better , another approach is ...

(1) Let e^x = t







This is factorable (if it wasn't you could use a quadratic formula)

Solve for t , then use (1) to solve for x , then check the solutions by substituting for x in your original problem.

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October 24th, 2013, 08:47 AM   #6
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Re: Another exponential equation....

Thanks guys!!!...I got it right once I wrote the the question correct
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