If log 3 = x, log 7 = y, log 10 = 1, and A = 3000/49, how do you find log A in terms of x and y?

1 Answer
Aug 11, 2016

Answer:

#logA=x-2y+3#.

Explanation:

#logA=log(3000/49)=log3000-log49..............[as, log(a/b)=loga-logb]#

#=log(3*10^3)-log7^2#

#=log3+log10^3-2log7..........[as, logab=loga+logb]#

#=log3+3log10-2log7............[as, loga^m=mloga]#

#=x+3-2y#

#logA=x-2y+3#.