Another maxwell bolltzman problem

[AStaunton]

want to show v_peak=sqrt(2kT/m):

the point here is that have a maximum when df/dv=0

max-boltz equation again is:

$$f(v)=4\pi(\frac{m}{2\pi mkT})^{\frac{3}{2}}v{}^{2}e^{-\frac{mv^{2}}{2kT}}$$

and taking derivitive and setting=0 and then dividing out any constant expressions gives:

$$\frac{df}{dv}=(2v)e^{-mv^{2}/2kT}-(2kT)e^{-mv^{2}/2kT}=0\implies e^{-mv^{2}/2kT}(2v-2kT)=0$$

I then divided out the (2v-2kT) term and took natural log of both sides:

$$-\frac{mv^{2}}{2kT}=1\implies v=\sqrt{\frac{-2kT}{m}}$$

clearly the minus sign should not be in there...besides that the answer is correct...can someone please tell me how I ended with an extra minus sign within the squareroot?ie.what mistake did I make do produce a minus sign along the way...

Another question about the above is:
I divided out the (2v-2kT/m) term and then proceeded to take logs and so on...however since this term also contains a v, I expected that I should if I wanted be able to divide out the exponential term instead, leaving me with:

$$2v-2kT/m=0$$

and again solving this equation for v should give me another way to find v_peak, however from this equation v=kT/m which is not the v_peak expression I wanted...can someone tell me why this is so?

 

[AStaunton]

****error update****

at some points I wrote 2kT where I should have written 2kT/m