Provelog_(1/p) x = -log_p (x) (x > 0). I understand that if I flip the base 1/p to p I'll get -log_p (x). But how do I prove it?

Hello!
We want to prove the identity  log_(1/p)(x) = -log_p(x)  (for all xgt0, pgt0 and p != 1 ).
To do this, let's recall what logarithms are. By the definition,  log_b(a) is such a number c that  b^c = a.  It is known that such a number always exists and is unique (of course for agt0, bgt0 and b != 1 ).
Therefore to verify this identity we raise 1/p to the power of each side:
(1/p)^(log_(1/p)(x)) = (1/p)^(-log_p(x))
(the function (1/p)^x is one-to one on its domain, therefore this operation gives an equivalent equality).
The left part is equal to x by the definition of logarithm, what about the right part? We'll use some properties of powers, 1/p=p^(-1) and (p^a)^b=p^(a*b):
(1/p)^(-log_p(x)) =(p^(-1))^(-log_p(x)) =p^((-1)*(-log_p(x))) =p^(log_p(x)).
And this is equal to x, too, by the definition of logarithm. This way we proved the desired identity.