# transposition of formula

• Mar 12th 2008, 01:39 PM
al2308
transposition of formula
I've got the following formula

X=k(1-e^(-t/T)

i've managed as part of my assignment to find values for all of the constants or variables except t.

so i now have the formula 10=20(1-e^(-t/0.1)

could anybody show me how to transpose this formula for t ?
• Mar 12th 2008, 02:11 PM
TheEmptySet
You bet...

First we need to isolate the term containing the variable.

$\displaystyle 10=20(1-e^{-t/.1})$

$\displaystyle \frac{1}{2}=1-e^{-t/.1}$

$\displaystyle -\frac{1}{2}=-e^{-t/.1}$

$\displaystyle \frac{1}{2}=e^{-t/.1}$

taking the natural log of both sides gives.

$\displaystyle ln(1/2)=ln(e^{-t/.1})$

$\displaystyle ln(1)-ln(2)= \frac{-t}{.1}ln(e)$

$\displaystyle -ln(2)=\frac{-t}{.1}$

solving for t gives...

$\displaystyle .1ln(2)=t$
• Mar 12th 2008, 02:27 PM
topsquark
Quote:

Originally Posted by al2308
I've got the following formula

X=k(1-e^(-t/T)

i've managed as part of my assignment to find values for all of the constants or variables except t.

so i now have the formula 10=20(1-e^(-t/0.1)

could anybody show me how to transpose this formula for t ?

$\displaystyle 10 = 20 \left ( 1 - e^{-t/0.1} \right )$

$\displaystyle \frac{1}{2} = 1 - e^{-t/0.1}$

$\displaystyle e^{-t/0.1} = \frac{1}{2}$

$\displaystyle -\frac{t}{0.1} = ln \left ( \frac{1}{2} \right ) = - ln(2)$

$\displaystyle t = 0.1 \cdot ln(2)$

-Dan