Prove`log_(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?

Expert Answers

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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.

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