# Re-arranging formula

• January 13th 2011, 03:31 AM
kenmunger
Re-arranging formula
Hi, im new here. Just wondering if someone could walk me through solving this for equation for V2. I have a solution but some parts seemed to have been skipped and i dont quite understand why:

150*V2^1.3 = 100*((5*10^-3)+V2)^1.3

Thanks
• January 13th 2011, 03:38 AM
e^(i*pi)
$150V_2^{1.3} = 100(V_2 + 5 \cdot 10^{-3})^{1.3}$

Take the log of both sides (I shall use natural log)

$\ln(150V_2^{1.3}) = \ln[100(V_2 + 5 \cdot 10^{-3})^{1.3}]$

You can then use the laws of logs to separate what's inside

$\ln(150) + 1.3\ln(V_2) = \ln(100) + 1.3\ln(V_2+5 \cdot 10^{-3})$

$1.3[\ln(V_2+5 \cdot 10^{-3}) - \ln(V_2)] = \ln(150) - \ln(100) = \ln(1.5)$

$\ln \left(\dfrac{V_2+5 \cdot 10^{-3}}{V_2}\right) = \dfrac{\ln(1.5)}{1.3}$

Chances are there is an easier way than using logs but there is more than one way to skin a mathematical cat :)
• January 13th 2011, 03:41 AM
Prove It
$\displaystyle 150V_2^{\frac{13}{10}} = 100(5\cdot 10^{-3} + V_2)^{\frac{13}{10}}$

$\displaystyle \frac{150}{100} = \frac{(5\cdot 10^{-3} + V_2)^{\frac{13}{10}}}{V_2^{\frac{13}{10}}}$

$\displaystyle \frac{3}{2} = \left(\frac{5\cdot 10^{-3} + V_2}{V_2}\right)^{\frac{13}{10}}$

$\displaystyle \left(\frac{3}{2}\right)^{\frac{10}{13}} = \frac{5\cdot 10^{-3} + V_2}{V_2}$

$\displaystyle \frac{3^{\frac{10}{13}}}{2^{\frac{10}{13}}} = \frac{5\cdot 10^{-3}}{V_2} + 1$

Can you go from here?
• January 13th 2011, 03:45 AM
kenmunger
I thought i could but im not getting the same answer as my solution.

Logs were never my strong point...
• January 13th 2011, 03:47 AM
Prove It
You don't need logarithms, see my post above...
• January 13th 2011, 04:02 AM
kenmunger
Yeah, i just worked through it compared to the given solution. It has been done the same way as you "pove it" only difference is your (3^10/13)/(2^10/13) is written as (150/100)^1/1.3. Thanks for the help both of you, i can make sense of it now :)
• January 13th 2011, 04:11 AM
kenmunger
I tell a lie... I'm not 100% sure why the +1 gets rid of the V2?!?
• January 13th 2011, 05:06 AM
DrSteve
Prove it split the fraction into two pieces. More precisely he used the following:

$\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$.

Do you see it now?
• January 13th 2011, 06:27 AM
kenmunger
I do... spot on that, cheers :)