# 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

Pratham Ashish
17 Points
15 years ago

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/x2 <1/x, hence

(x-1) / x <y

Pratham Ashish
17 Points
15 years ago
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