- #1

Stoney Pete

- 49

- 1

^{a}log x =

^{b}log x /

^{b}log a

For all the all other rules concerning logarithms I could derive them from the rules for exponentiation, except for this rule. Could someone please explain to me how this derivation works? And how the value of be can be practically determined so that it gives a neat answer for

^{b}log a.

For example as applied to

^{3/2}log 3 =

^{b}log 3 /

^{b}log 3/2

How to determine the value of b in this case such that

^{b}log 3/2 gives a neat answer?

Thanks in advance!