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
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
$\displaystyle 150V_2^{1.3} = 100(V_2 + 5 \cdot 10^{-3})^{1.3}$
Take the log of both sides (I shall use natural log)
$\displaystyle \ln(150V_2^{1.3}) = \ln[100(V_2 + 5 \cdot 10^{-3})^{1.3}]$
You can then use the laws of logs to separate what's inside
$\displaystyle \ln(150) + 1.3\ln(V_2) = \ln(100) + 1.3\ln(V_2+5 \cdot 10^{-3})$
$\displaystyle 1.3[\ln(V_2+5 \cdot 10^{-3}) - \ln(V_2)] = \ln(150) - \ln(100) = \ln(1.5)$
$\displaystyle \ln \left(\dfrac{V_2+5 \cdot 10^{-3}}{V_2}\right) = \dfrac{\ln(1.5)}{1.3}$
Chances are there is an easier way than using logs but there is more than one way to skin a mathematical cat
$\displaystyle \displaystyle 150V_2^{\frac{13}{10}} = 100(5\cdot 10^{-3} + V_2)^{\frac{13}{10}}$
$\displaystyle \displaystyle \frac{150}{100} = \frac{(5\cdot 10^{-3} + V_2)^{\frac{13}{10}}}{V_2^{\frac{13}{10}}}$
$\displaystyle \displaystyle \frac{3}{2} = \left(\frac{5\cdot 10^{-3} + V_2}{V_2}\right)^{\frac{13}{10}}$
$\displaystyle \displaystyle \left(\frac{3}{2}\right)^{\frac{10}{13}} = \frac{5\cdot 10^{-3} + V_2}{V_2}$
$\displaystyle \displaystyle \frac{3^{\frac{10}{13}}}{2^{\frac{10}{13}}} = \frac{5\cdot 10^{-3}}{V_2} + 1$
Can you go from here?