Question

Show by using mean value theorem and taking f(x)=logx that 1-a/b < logb/a < b/a -1

Answer 1

f(x)=log(x)
so f'(x)=1/x.

By Meaan Value Theorem,

f'(c) = (f(b)-f(a))/(b-a)
= (log b - log a)/(b-a)
= log(b/a)/(b-a)=1/c where c lies between a and b.

Now, since 1/x is a decreasing function, and a < c < b,
we get 1/b < 1/c < 1/a
1/b<log(b/a)/(b-a)<1/a

multiply (b-a)
(b-a)/b<log(b/a)<(b-a)/a
(1 - a/b)<log(b/a)<(b/a - 1)