# Isolate $V$ in $\frac1b\ln(\frac V{V_0}) - \frac cb \ln \frac{b+cV}{b+cV_0} = -\frac tm$

I have to isolate $$V$$ in this equation. I have the final result in my book, but I'm not even close to get it.

$$\frac1b\ln\left(\frac V{V_0}\right) - \frac cb \ln \frac{b+cV}{b+cV_0} = -\frac tm$$

I tried a lot of things, but none of them works. I'm pretty sure I did something wrong with ln.

$$\frac1b\ln\left(\frac {\frac V{V_0}}{\frac{b+cV}{b+cV_0}}\right)(1-c) = -\frac tm$$

or

$$\frac {1-c}b\ln\left(\frac {\frac V{V_0}}{\frac{b+cV}{b+cV_0}}\right) = -\frac tm$$

I'm not sure how to manipulate ln.

• Doesn't seem possible. At least Maple cannot solve it and gives the solution in terms of the "RootOf" function, which means that it is probably not possible in closed form.
– user127032
Commented Sep 19, 2020 at 0:58
• Alright, so basically I muffed the integral. Commented Sep 19, 2020 at 1:06
• Hiding the full context makes users (willing to help you) wasting thei time. Commented Sep 19, 2020 at 2:55

From $$\frac{1-c}{b}\ln\left(\frac{\frac{V}{V_0}}{\frac{b+cV}{b+cV_0}}\right)=\frac{1-c}{b}\ln\left(\frac{V(b+cV_0)}{V_0(b+cV)}\right)=-\frac{t}{m}$$, it follows that $$\ln\left(\frac{V(b+cV_0)}{V_0(b+cV)}\right)=-\frac{bt}{(1-c)m}=\frac{bt}{(c-1)m}$$. Since we know that $$e^{\ln(x)}=x$$, we obtain $$\frac{V(b+cV_0)}{V_0(b+cV)} = e^{\frac{bt}{(c-1)m}}$$ from which you should be able to solve it for $$V$$.

• That's what I had found, but I can't isole V. That's why I thought I did something wrong. Commented Sep 19, 2020 at 1:00
• No worries. Denote $x=e^{stuff}$. Then we have $\frac{V(b+cV_0)}{V_0(b+cV)} = x \Leftrightarrow V(b+cV_0) = xV_0(b+cV) = bxV_0 + cxV_0V \Leftrightarrow V(b+cV_0-cxV_0) = V(b+c(1-x)V_0) = bxV_0 \Leftrightarrow V = \frac{bxV_0}{b+c(1-x)V_0}$. Assuming no miscalculations.
– user826493
Commented Sep 19, 2020 at 1:10
• Alright thanks. However, I muffed the integral. $\int_{0}^{t} dt = -m \int_{V0}^{v} \frac {dv}{bv+cv^2}$ Commented Sep 19, 2020 at 1:50

$$\int_0^t dt = -\frac m{-1} \int_{V0}^v \frac {dv}{bv+cv^2}$$

My final answer is $$v = \frac{e^{\frac{-bt}{cm}} VoB}{b +cVo- e^{\frac{-bt}{cm}}VoC}$$

The right answer is $$v = \frac{b +cVo}{b +cVo- e^{\frac{-bt}{m}}VoC} -1$$

$$\frac1b\ln\frac V{V_0} - \frac cb \ln \frac{b+cV}{b+cV_0} = -\frac tm$$

These two terms have the factor $$1/b$$ in common, so you get $$\frac1b \left( \ln\frac V{V_0} - c \ln \frac{b+cV}{b+cV_0}\right) = -\frac tm$$ But you cannot pull $$1-c$$ out of that because the things by which the $$1$$ is multiplied is not the same as the thing by which the $$c$$ is multiplied.

But you can write $$\ln\left(\left( \frac V{V_0} \right)^{1/b} \right) - \ln\left( \left( \frac{b+cV}{b+cV_0} \right)^{c/b} \right) = -\frac t m$$ and then $$\ln \left( \frac{(V/V_0)^{1/b}}{\left( \frac{b+cV}{b+cV_0} \right)^{c/b}} \right) = -\frac t m$$ $$\frac{(V/V_0)^{1/b}}{\left( \frac{b+cV}{b+cV_0} \right)^{c/b}} = e^{-t/m}$$ $$\frac{(V/V_0)}{\left( \frac{b+cV}{b+cV_0} \right)^c} = e^{-bt/m}$$ $$\frac{V}{V_0}\cdot \left( \frac{b+cV}{b+cV_0} \right)^c = e^{-bt/m}$$ $$V(b+cV)^c = e^{-bt/m} \cdot V_0(b+cV_0)^c$$ $$V^{1/c} (b+cV) = e^{-bt/(cm)} \cdot V_0^{1/c} (b+cV_0)$$

To solve this for $$V,$$ you probably need numerical methods.