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# Prove that

Harsh Patodia IIT Roorkee
askIITians Faculty 907 Points
6 years ago
Hi student

This is of the type  f(x)g(x) which is of form 1 such that f(x)1 and g(x)

Solution of such form is give by e g(x)(f(x)-1)

When you will subsitute appropriate values of f(x) and g(x) you can get the answer.
Shibashis Mallik
21 Points
6 years ago
Is it necessary to subtract 1 from f(x) in the form eg(x)(f(x)-1)  ?
Harsh Patodia IIT Roorkee
askIITians Faculty 907 Points
6 years ago
Hi Student,

Basically subtraction of 1 comes from the derivation. So it must be done.
If you want to know the derivation please reply on this thread otherwise its not necessary. You can remember the form the apply this directly.
Shibashis Mallik
21 Points
6 years ago
Yes, why not? I definitely want to know the derivation.
Harsh Patodia IIT Roorkee
askIITians Faculty 907 Points
6 years ago
let y = f(x)g(x)
Taking log on both sides
log y= g(x) log(f(x))
Taking antilog on both sides
y = eg(x) logf(x)

Consider log f(x)= log ( 1 + (f(x)-1))
Expansion of log ( 1 + x)= x – x2/2 + x3/3 . . . .
log ( 1+ (f(x)-1)) = (f(x) – 1) – (f(x)-1)2/2 . . . . . .
since f(x) is tending to f(x) – 1 will be very small so higher are neglected
this means log ( 1+ (f(x)-1) = f(x) – 1 if f(x) tends to 1
The above expression becomes
y= eg(x) ( f(x) -1)