Question 35550
Let y=log(baseA)B
Then A^y=B
Take the log(baseC) of both sides to get:
log(baseC)A^y=log(baseC)B
Using the law of logs which says logA^n=nlogA you get:
y(log(baseC)A = log(baseC)B
y= [log(baseC)B/[log(baseC)A}
Therefore log(baseA)B = [log(baseC)B]/[log(baseC)A]
Cheers,
Stan H.