Hi Guys! algebra help needed for engineering studies!

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February 18th 2013, 02:42 PM
andym1512

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!)
₂The next question is

A= V (Q- mV²) Where Q = 50.28, m = 17, V = 5 and g = 9.81
100 g

Many thanks
Andy
February 18th 2013, 02:48 PM
andym1512

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
February 18th 2013, 02:54 PM
jakncoke

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

$16 = 24(1-e^{-\frac{t}{2}})$
divide both sides by 24 to get
$\frac{2}{3} = 1 -e^{-\frac{t}{2}}$
add - 1 to both sides
$\frac{-1}{3} = -e^{-\frac{t}{2}}$
multiply both sides by -1
$\frac{1}{3} = e^{-\frac{t}{2}})$
take natural log of both sides
$ln(\frac{1}{3}) = -\frac{t}{2}$
use log rules $a*ln(x) = ln(x^a)$
to get $2*ln(\frac{1}{3}) = ln((\frac{1}{3})^2) = ln(\frac{1}{9})$
so $-t = ln(\frac{1}{9})$
$t = - ln(\frac{1}{9})=ln((\frac{1}{9})^{-1}) = ln(9)$
t = ln(9)
February 18th 2013, 02:55 PM
andym1512

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

thanks very much for reply, really helped!
February 20th 2013, 04:24 PM
LeonidaD

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

Quote:

Originally Posted by jakncoke

$16 = 24(1-e^{-\frac{t}{2}})$
divide both sides by 24 to get
$\frac{2}{3} = 1 -e^{-\frac{t}{2}}$
add - 1 to both sides
$\frac{-1}{3} = -e^{-\frac{t}{2}}$
multiply both sides by -1
$\frac{1}{3} = e^{-\frac{t}{2}})$
take natural log of both sides
$ln(\frac{1}{3}) = -\frac{t}{2}$
use log rules $a*ln(x) = ln(x^a)$
to get $2*ln(\frac{1}{3}) = ln((\frac{1}{3})^2) = ln(\frac{1}{9})$
so $-t = ln(\frac{1}{9})$
$t = - ln(\frac{1}{9})=ln((\frac{1}{9})^{-1}) = ln(9)$
t = ln(9)

you are a genius..