Hi Guys! algebra help needed for engineering studies!

Hi there Im andy from Liverpool

im currently studying for an engineering degree, but firstly have to complete a foundation course in maths. Its going well so far - until the dreaded algebra questions came along! Ive understood some of it so far but the assesment questions have tied me in knots and i feel completely stuck. These are the questions that are giving me particular difficulty, any help would be much appreciated.Solve: 16 = 24(1-e^{-t}) (It shgould read -t over 2 but cant figure out to write correctly on here!)

_{2 }The next question is

A= __ V __ (Q- __mV__^{2}) Where Q = 50.28, m = 17, V = 5 and g = 9.81

100 g

Many thanks

Andy

Re: Hi Guys! algebra help needed for engineering studies!

Sorry second equation isnt very clear should be V divided by 100 mV squared divided by g

Re: Hi Guys! algebra help needed for engineering studies!

$\displaystyle 16 = 24(1-e^{-\frac{t}{2}}) $

divide both sides by 24 to get

$\displaystyle \frac{2}{3} = 1 -e^{-\frac{t}{2}} $

add - 1 to both sides

$\displaystyle \frac{-1}{3} = -e^{-\frac{t}{2}}$

multiply both sides by -1

$\displaystyle \frac{1}{3} = e^{-\frac{t}{2}})$

take natural log of both sides

$\displaystyle ln(\frac{1}{3}) = -\frac{t}{2}$

use log rules $\displaystyle a*ln(x) = ln(x^a) $

to get $\displaystyle 2*ln(\frac{1}{3}) = ln((\frac{1}{3})^2) = ln(\frac{1}{9})$

so $\displaystyle -t = ln(\frac{1}{9})$

$\displaystyle t = - ln(\frac{1}{9})=ln((\frac{1}{9})^{-1}) = ln(9)$

t = ln(9)

Re: Hi Guys! algebra help needed for engineering studies!

thanks very much for reply, really helped!

Re: Hi Guys! algebra help needed for engineering studies!

Quote:

Originally Posted by

**jakncoke** $\displaystyle 16 = 24(1-e^{-\frac{t}{2}}) $

divide both sides by 24 to get

$\displaystyle \frac{2}{3} = 1 -e^{-\frac{t}{2}} $

add - 1 to both sides

$\displaystyle \frac{-1}{3} = -e^{-\frac{t}{2}}$

multiply both sides by -1

$\displaystyle \frac{1}{3} = e^{-\frac{t}{2}})$

take natural log of both sides

$\displaystyle ln(\frac{1}{3}) = -\frac{t}{2}$

use log rules $\displaystyle a*ln(x) = ln(x^a) $

to get $\displaystyle 2*ln(\frac{1}{3}) = ln((\frac{1}{3})^2) = ln(\frac{1}{9})$

so $\displaystyle -t = ln(\frac{1}{9})$

$\displaystyle t = - ln(\frac{1}{9})=ln((\frac{1}{9})^{-1}) = ln(9)$

t = ln(9)

you are a genius..