# Making x the subject in an exponential equation

• Jan 8th 2010, 09:53 PM
smmmc
Making 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 PM
Prove It
Quote:

Originally Posted by smmmc
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 ??

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

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

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

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

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

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

(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 PM
smmmc
• Jan 8th 2010, 10:11 PM
Sudharaka
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 PM
mr fantastic
Quote:

Originally Posted by smmmc

Which is one of the forms of the answer given by Prove It. And you have been told in another post what ln means. So what's your problem here?
• Jan 9th 2010, 04:28 AM
HallsofIvy
Quote:

Originally Posted by smmmc
make x the subject
y=5(1-e^(-x))

first i expand it

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

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

This is wrong. log(a+ b) is not equal to log(a)+ log(b)

As others have said, the simplest thing to do is write y- 5= -5e^{-x+} and take the logarithm of that.
loge(y)-loge(5)=5x

Quote:

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