y-logz

[nadinex]

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-log(zx10^-y) = y-logz

do I have the expression down right?
I don't think so? Because log of 1x10^-3 ---> with the equation, comes out to be 3 - log (1) = 3
(not -3)

 

[Cawolf]

-log (z * 10^-y)

= -[log(z) + log(10^-y)]

= -log(z) - log(10^-y)

=-log(z) -(-y)

=y - log(z)

So yes, you are right.

In your example with 1 * 10^-3 you did not take the negative log, so it isn't really the same.