SOLUTION: prove that log15(1+(log30)/(log15))+1/2(log16(1+(log7)/(log4)))-log6((log3)/(log6)+1+(log7)/(log6))=2

Question 1016891: prove that log15(1+(log30)/(log15))+1/2(log16(1+(log7)/(log4)))-log6((log3)/(log6)+1+(log7)/(log6))=2
Answer by mathmate(429) (Show Source): You can put this solution on YOUR website!

Question:
prove that
log15(1+(log30)/(log15))+1/2(log16(1+(log7)/(log4)))-log6((log3)/(log6)+1+(log7)/(log6))=2

Solution:

We used the symbol E to stand for the value of the left-hand side, or
E=log15(1+(log30)/(log15))+1/2(log16(1+(log7)/(log4)))-log6((log3)/(log6)+1+(log7)/(log6))

We need to use some laws of logarithm:
log(a)+log(b)=log(ab)
log(a)-log(b)=log(a/b)
(1/2)log(a)=log(sqrt(a))

Also, note that log(x) implies log(x) to the base 10.

The idea of solution is to break up each term and simplify.
We have:

log15(1+(log30)/(log15))=log(15)+log(30)=log(450).........(1)

1/2(log16(1+(log7)/(log4)))=log(sqrt(16))(1+log(7)/log(4))
=log(4)(1+log(7)/log(4))
=log(4)+log(7)
=log(28).................................................(2)

-log6((log3)/(log6)+1+(log7)/(log6))
=-(log(3)+log(6)+log(7)
=-(log(3*6*7))
=-log(126)...............................................(3)

Putting the three terms together we have:

E=E=log15(1+(log30)/(log15))+1/2(log16(1+(log7)/(log4)))-log6((log3)/(log6)+1+(log7)/(log6))
=log(450)+log(28)-log(126)
=log(450*28/126)
=log(100)
=log(10^2)
=2 which is exactly the value of the right hand side, so proved.

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