Question 1016891
 
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.