Making x the subject in an exponential equation

Printable View

January 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 ??
January 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 ??

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

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

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

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

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

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

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

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

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

$x = \ln{5} - \ln{(5 - y)}$ .
January 8th 2010, 10:07 PM
smmmc

the answer says loge(5/5-y) ?
January 8th 2010, 10:11 PM
Sudharaka

Dear smmmc,

Of course, $\ln\frac{5}{5-y}=\ln5-\ln(5-y)$ Therefore what Prove it had shown is the answer.
January 8th 2010, 10:12 PM
mr fantastic

Quote:

Originally Posted by smmmc

the answer says loge(5/5-y) ?

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?
January 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 ??