Question

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?

Answer 1

`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`.