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

[sphyics]

f(x)= lnx ; g(x) = e^x

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

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

how to solve the problem algebraically.

 

[Hootenanny]

f(x)= lnx ; g(x) = e^x

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

g{f(x)}= e^{ln 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:

f(x) and g(x) are inverse of each other.

 

[sphyics]

You answered the question yourself:

g{f(x)}= e^lnx = ? how to solve this equation algebraically and come to a solution..

 

[micromass]

What does it mean that f and g are inverses of each other??

 

[Number Nine]

g{f(x)}= e^lnx = ? 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.

Spoiler

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

 

[sphyics]

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.

Spoiler

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

y=x=e^{\ln x}

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 :)

 

[sphyics]

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.

 

[paulfr]

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