Apply algebraic reasoning to show that a=b^(loga/logb) for any a,b>0

Answer 1

take the log of each side
loga=(loga/logb)logb
now reduce.

Answer 2

I have:

loga=(loga/logb)logb
(loga/logb )(1/loga)=logb
(1/logb)=logb

Ok now what?

Answer 3

you erred.
loga=(loga/logb)logb
the logb on the right side divide out (one on numerator, one in denominator)
loga=loga
divide both sides by loga.