Solve Exponentials Problem: ln((V+v)/(V-v)) = 2ctV

[Nevermore]

I have ln((V+v)/(V-v)) = 2ctV, and I've checked this is right. But how do I go from here to v = V((e^Vct - e^-Vct)/(e^Vct+e^-Vct))? The solutions just give them as successive lines.

(I appreciate that's not the easiest thing to read, it's taken from the 2004 MEI specimen paper for A-level Mechanics 4, if that helps.)

 

[Hootenanny]

HINT: Laws of logs;

$$\ln\frac{a}{b} = \ln|a| - \ln|b|$$

 

[Nevermore]

OK, so
ln((V+v)/(V-v)) = 2ctV
=> ln(V+v)-ln(V-v) = 2ctV
ln(V)ln(v) - ln(V)/ln(v)
2ln(v) = 2ctV
v = e^Vct

Is this right? Where can I go from here?

 

[geniusprahar_21]

OK, so
ln((V+v)/(V-v)) = 2ctV
=> ln(V+v)-ln(V-v) = 2ctV
ln(V)ln(v) - ln(V)/ln(v)
2ln(v) = 2ctV
v = e^Vct

Is this right? Where can I go from here?

there is something wrong in what u have written... first of all
log(a*b) = log(a) + log(b)...what u have used is log(a+b) = log(a) * log(b)..even after that u have written something wrong...check it once more...
anyway u don't need to use that...use the basic definition of logs...
if ln(a) = b
=> e^b = a

 

[HallsofIvy]

In other words,
$$\frac{V+v}{V-v}= e^{2ctV}$$
Solve that for v.