make x the subject

y=5(1-e^(-x))

first i expand it

so

y=5-5e^(-x)

loge(y)=loge(5)+loge(e5x)

loge(y)-loge(5)=5x

x=loge(y)-loge(5)/5 ??

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- Jan 8th 2010, 09:53 PMsmmmcMaking x the subject in an exponential equation
make x the subject

y=5(1-e^(-x))

first i expand it

so

y=5-5e^(-x)

loge(y)=loge(5)+loge(e5x)

loge(y)-loge(5)=5x

x=loge(y)-loge(5)/5 ?? - Jan 8th 2010, 09:58 PMProve It
$\displaystyle y = 5\left(1 - e^{-x}\right)$

$\displaystyle \frac{y}{5} = 1 - e^{-x}$

$\displaystyle e^{-x} = 1 - \frac{y}{5}$

$\displaystyle e^{-x} = \frac{5 - y}{5}$

$\displaystyle -x = \ln{\frac{5 - y}{5}}$

$\displaystyle x = -\ln{\frac{5 - y}{5}}$

(Anything from this step onwards is just to get the answer to whatever form you like)

$\displaystyle x = \ln{\left(\frac{5 - y}{5}\right)^{-1}}$

$\displaystyle x = \ln{\frac{5}{5 - y}}$

$\displaystyle x = \ln{5} - \ln{(5 - y)}$. - Jan 8th 2010, 10:07 PMsmmmc
the answer says loge(5/5-y) ?

- Jan 8th 2010, 10:11 PMSudharaka
Dear smmmc,

Of course,$\displaystyle \ln\frac{5}{5-y}=\ln5-\ln(5-y)$ Therefore what Prove it had shown is the answer. - Jan 8th 2010, 10:12 PMmr fantastic
- Jan 9th 2010, 04:28 AMHallsofIvy