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Mar 2010
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Hi guys. I'm stuck on this one.

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

Any help would be appreciated.
 

masters

MHF Helper
Jan 2008
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1,187
Big Stone Gap, Virginia
Hi guys. I'm stuck on this one.

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

Any help would be appreciated.
Hi Alvy,

Do this:

\(\displaystyle \ln 2e^{-x}=\ln 3e^{.1x}\)

\(\displaystyle \ln 2 + (-x)=\ln 3 + .1x\)

\(\displaystyle -1.1x=\ln 3 - \ln 2\)

\(\displaystyle x=\frac{\ln 3 - \ln 2}{-1.1}\)
 
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Soroban

MHF Hall of Honor
May 2006
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Lexington, MA (USA)
Hello, Alvy!

Another approach . . .


Solve for \(\displaystyle x\!:\;\;2e^{-x} \:=\: 3e^{0.1x}\)

We have: . \(\displaystyle 3e^{0.1x} \;=\;2e^{-x}\)

Multiply by \(\displaystyle \frac{e^x}{3}\!:\;\;e^{1.1x} \;=\;\frac{2}{3}\)

\(\displaystyle \text{Take logs: }\;\ln\left(e^{1.1x}\right) \;=\;\ln\left(\frac{2}{3}\right) \quad\Rightarrow\quad 1.1x\underbrace{\ln(e)}_{\text{This is 1}} \;=\;\ln\left(\frac{2}{3}\right) \)


Therefore: . \(\displaystyle 1.1x \;=\;\ln\left(\frac{2}{3}\right) \quad\Rightarrow\quad x \;=\;\frac{1}{1.1}\ln\left(\frac{2}{3}\right) \)