Asked by Jus

Apply algebraic reasoning to show that a=b^(loga/logb) for any a,b>0
Answered by bobpursley
take the log of each side
loga=(loga/logb)logb
now reduce.
Answered by Jus
I have:

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

Ok now what?
Answered by bobpursley
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.
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