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Composite function help please. :)

Hello there,

Cold someone tell me how I would work out fg where f(x) = 3e^2x and g(x)=ln4x. I know how to form a composite function i'm just stuck on what to do as there is and e and ln in the equation. Thanks! :smile:
The same way as you usually would. What do you get when you do it? Post your working. Also, is that exp(2x) or exp^2(x)?
Reply 2
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apo1324
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Original post by little_wizard123
The same way as you usually would. What do you get when you do it? Post your working. Also, is that exp(2x) or exp^2(x)?


Hi, it's exp^2x, sorry.

My working is:-

(3)(e^(2)(ln4x)) = fg. Is this correct? Is this all I can do? Thanks!
You can use a log rule to simplify it. And then use the fact that ln(x) and e^x are inverses of each other.
Original post by apo1324
Hi, it's exp^2x, sorry.

My working is:-

(3)(e^(2)(ln4x)) = fg. Is this correct? Is this all I can do? Thanks!


If you mean 3e2ln⁡(4x) 3e^{2\ln{(4x)}} , then that's right.

You can simplify it more though. Note that
Unparseable latex formula:

e^{\ln{(\mbox{stuff})}} = \mbox{stuff}

Reply 5
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apo1324
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Original post by Daniel Freedman
If you mean 3e2ln⁡(4x) 3e^{2\ln{(4x)}} , then that's right.

You can simplify it more though. Note that
Unparseable latex formula:

e^{\ln{(\mbox{stuff})}} = \mbox{stuff}



Is the answer just 12x? Thanks. :smile: Also, if the domain of f was (-infinite,infinite), and the domain of g was (0,infinite):-

Would the domain of fg be [0,infinite) and the range of fg be [0,infinite)?? I don't know if this is correct.
Original post by apo1324
Is the answer just 12x? Thanks. :smile: Also, if the domain of f was (-infinite,infinite), and the domain of g was (0,infinite):-

Would the domain of fg be [0,infinite) and the range of fg be [0,infinite)?? I don't know if this is correct.


Nope. You can't use exp (ln(x)) = x if there is a number in front of the log. You need to use the rule that a*ln(x) = ln(x^a) first.
Reply 7
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apo1324
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Original post by little_wizard123
Nope. You can't use exp (ln(x)) = x if there is a number in front of the log. You need to use the rule that a*ln(x) = ln(x^a) first.


Is it this:-

3e^ln4x^2? :s-smilie:
Original post by apo1324
Is it this:-

3e^ln4x^2? :s-smilie:


(4x)^2 which is 16x^2.

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