# Thread: Making x the subject in an exponential equation

1. ## 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 ??

2. 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)

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

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

$\displaystyle x = \ln{5} - \ln{(5 - y)}$.

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

4. Dear smmmc,

Of course,$\displaystyle \ln\frac{5}{5-y}=\ln5-\ln(5-y)$ Therefore what Prove it had shown is the answer.

5. 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?

6. 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

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