If y=log _e x then for x>1, show that we must have

(a) x-1>y (b) x²-1 >y (c) (x-1) / x

 hi,

if two functions are equal at a point , & after that point if one has always a high derivative than other , it means that it increases  more with the same change in x than other,it clearly emplies that it would always be high than other

(a) x-1>y 

   at x =1 ,

l.h.s.=r.h.s . = 0

d/dx(l.h.s) = d/dx(x-1) =1

d/dx(r.h.s) = d/dx (log _e x) = 1/x

if x >1,

1/x <1

d/dx(l.h.s) > d/dx(r.h.s)

emplies,

x-1>y 

 

(b) x²-1 >y  ,                at x=1

l.h.s = rh.s = 0

d/dx(l.h.s) =d/dx (x²-1)

                = 2x

d/dx(r.h.s) =d/dx (log _e x ) =  1/x

we know,

       for x>1 ,

2x >1/x 

hence,

 x²-1 >y   

 

 (c)  (x-1) / x <y

at x= 1,

l.h.s = rh.s = 0,

d/dx(l.h.s) =d/dx ( (x-1) / x )

                = 1/x²

d/dx(l.h.s) =d/dx (log _e x )

                 = 1/x

for x>1 ,

  1/x^{2 _<}1/x, hence

 (x-1) / x <y

hi tushar, this problem is quite straight forward,,just start with writing x = e^y , now for 1st case substitute e^y in place of x and u will get (e^y - 1) > y, here u can see that the function on the left hand side is a exponential function and for x>1, e^y > 0 ,also for any value of y , the left side fxn will increase with a slope of e^y while right side fxn will increase only with a slope of 1 so, for all value of y ,(e^y - 1)>y........ similarly for 2nd case the eqn is (e^2y - 1)> y ,we can say that it is always valid for x>1...... now in 3rd case , ur eqn wud be (1 - 1/e^y) here u can see its slope is 1/e^y , so its slope is always less than 1 for all x>1....so the given condition (x-1) / x