SOLUTION: Please help me with this problem: logb^2 = .25596, logb^3 = .40568, logb^5 = .59432 logb^(5/12)^3 I know the answer is .096984, but don't know how to go about it. Thank

Question 20603: Please help me with this problem:
logb^2 = .25596, logb^3 = .40568, logb^5 = .59432
logb^(5/12)^3
I know the answer is .096984, but don't know how to go about it.
Thanks,
Sandy
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
logb^2 = .25596, logb^3 = .40568, logb^5 = .59432
this is not properly written..i think by the above you meant log 2 to base b =0.25596..etc....then the above has some meaning..otherwise they are not correct and represent contradictory things
logb^(5/12)^3=?...as given above this should mean that we want log (5/12)^3 to base b.
use the following formulae in logs which are true for any base..here wew can take b as base
log x*y=log x+log y
log x/y=log x-log y
log x^n=n*log x
log x to base b=log x /log b
LET US TAKE ALL LOGS TO BASE b AS WE HAVE THEM TO BASE b IN THE PROBLEM.
log(5/12)^3=3*log(5/12)
=3*(log 5-log 12)=3*(log 5-log 2*2*3)
=3*(log 5-log 2-log 2-log 3)=3*(0.59432-0.25596-0.25596-0.40568)
3*(-0.3238)=-0.96984
YOU GAVE THE ANSWER AS 0.096984...IN WHICH CASE THE QUESTION SHOULD BE log(5/12)^(-0.3) to base b =????.PLEASE CHECK BACK THE QUESTION WRITE IT DOWN PROPERLY AS AS GIVEN IN THE TEXT AND COME BACK

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