Derivatives: Solving for t -Xe^(-xt) + Ye^(-yt)

mrooney

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Oct 4, 2016
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I have this equation and I know what the final answer should be but I am stuck on how to get to it using derivatives. Thanks in advance
Solving for t

-Xe-xt + Ye-yt


Final answer should be
t= ln(y/x)(1/(y-x))
 
I have this equation and I know what the final answer should be but I am stuck on how to get to it using derivatives. Thanks in advance
Solving for t

-Xe-xt + Ye-yt


Final answer should be
t= ln(y/x)(1/(y-x))

That is not an equation - there is no "equal to"(=) sign in there.
 
I have this equation and I know what the final answer should be but I am stuck on how to get to it using derivatives. Thanks in advance
Solving for t

-Xe-xt + Ye-yt


Final answer should be
t= ln(y/x)(1/(y-x))

Assuming x and X are meant to be the same thing (and that's a BIG assumption) and the same for the Y and y:

\(\displaystyle -xe^{-xt} + ye^{-yt} =0 \)

\(\displaystyle y*e^{-yt} = x* e^{-xt} \)

\(\displaystyle \frac {y}{x} = \frac {e^{-xt}}{e^{-yt}} \)

\(\displaystyle \frac{y}{x} = e^{yt-xt}\)

Now take natural log of both sides and see if you can make t the subject.
 
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