Help with "e" function solving for x-logs?

drbeemanfromknecht

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Nov 5, 2013
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this is the problem: e^x-8e^-x=2.

what i have done is put everything in terms of base 2 to get

(2^0)(e^x)-(2^3)(e^-x)=2^1

from here, i am lost. i know i should factor out a two, but dont know what to do with the new coefficients left after taking out the 2. help?!
 
this is the problem: e^x-8e^-x=2.
what i have done is put everything in terms of base 2 to get
(2^0)(e^x)-(2^3)(e^-x)=2^1
\(\displaystyle \\{e^x} - 8{e^{ - x}} = 2\\{e^{2x}-2{e^x}}-8=0\)

Let \(\displaystyle y = {e^x}\), solve \(\displaystyle y^2-2y-8-0\).
 
Last edited:
this is the problem: e^x-8e^-x=2.

what i have done is put everything in terms of base 2 to get

(2^0)(e^x)-(2^3)(e^-x)=2^1

from here, i am lost. i know i should factor out a two, but dont know what to do with the new coefficients left after taking out the 2. help?!

e^x-8e^-x=2

\(\displaystyle \displaystyle e^x \ - \ 8 * \dfrac {1}{e^{x}} \ = \ 2\)

\(\displaystyle \displaystyle e^{2x} \ - \ 8 \ = \ 2 * e^x\)

substitute

u = ex

\(\displaystyle \displaystyle \left (e^{x}\right )^2 \ - \ 8 \ = \ 2 * e^x\)

\(\displaystyle \displaystyle \left (u\right )^2 \ - \ 8 \ = \ 2 * u\)

above is a quadratic equation - solve for 'u' - then back-substitute to solve for 'x'
 
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