## logarithm differentiation problem [closed]

I am reading a text book and am stumped with an example problem.

 y = n/p + 2*log(p)


First, y is differentiated with respect to p to give the following result:

 dy/dp  = -n/p^2 + 2/p


This is simple enough. Although I'm fairly certain you must assume the logarithm to the base e , i.e. is ln the natural logarithm.

The problem then details that you must solve dy/dp = 0 to find p which comes out as p = n/2

The last bit is confusing me, as p is then substituted back into y and the answer is simply given as 2log(n).

substituting the value for p into y, surely this implies:

2 + 2*log(n) - 2*log(2) = 2*log(n)


which implies than the logarithm is actually of base '2' not e as suggested by the original part of the problem. Am I going crazy? or can you change the base like this willy nilly. Has the text book made a mistake here?

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If $y$ is a solution, then so is $y+c$ for any constant $c$. Now observe that $2+2∗\log(n/2)$ and $2∗\log(n)$ differ by a constant. BTW this question is not appropriate here, it is not of research level. – GH Apr 19 2011 at 16:54