Solve Problem Algebraically: f{g(x)} and g{f(x)}

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sphyics
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f(x)= lnx ; g(x) = ex

f{g(x)} = ln ex = x; not an issue

g{f(x)}= eln x = ? (answer for this) f(x) and g(x) are inverse of each other.

how to solve the problem algebraically.
 
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sphyics said:
f(x)= lnx ; g(x) = ex

f{g(x)} = ln ex = x; not an issue

g{f(x)}= eln x = ? (answer for this) f(x) and g(x) are inverse of each other.

how to solve the problem algebraically.
You answered the question yourself:
sphyics said:
f(x) and g(x) are inverse of each other.
 
Hootenanny said:
You answered the question yourself:

g{f(x)}= elnx = ? how to solve this equation algebraically and come to a solution..
 
sphyics said:
g{f(x)}= elnx = ? how to solve this equation algebraically and come to a solution..

I suggest trying to prove it for yourself first. If you really can't, I've "spoilered" a proof below. It sounds like you may want to go back and brush up on some of your fundamentals.

[tex]y = e^{\ln x}[/tex]
[tex]\ln y = \ln e^{\ln x}[/tex]
[tex]\ln y = \ln x[/tex] (by the power rule of exponential functions, since ln e = 1)
[tex]y=x=e^{\ln x}[/tex]
 
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Number Nine said:
I suggest trying to prove it for yourself first. If you really can't, I've "spoilered" a proof below. It sounds like you may want to go back and brush up on some of your fundamentals.

[tex]y = e^{\ln x}[/tex]
[tex]\ln y = \ln e^{\ln x}[/tex]
[tex]\ln y = \ln x[/tex] (by the power rule of exponential functions, since ln e = 1)

[tex]y=x=e^{\ln x}[/tex]

OMG i was perfect till ln y = ln x; after that i confused myself over if ln y = ln x; does that imply y = x, now i can see the perfect picture, thanks very much for shedding light over the darkness my ignorance :)
 
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micromass said:
What does it mean that f and g are inverses of each other??

g{f(x)} = f{g(x)}, and both are one to one, hence inverse of each other.
 
Algebraic Solution ??
OK
Let the solution be called "c"
e^(Ln x) = c
Take the Ln of both sides
Ln [ e^(Ln x) ] = Ln c
Use the Power Rule for Logs to get
Ln x Ln e = Ln c
but Ln e = 1 so
Ln x = Ln c
thus
c = x
 
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