The Student Room Group

Natural log question?

The function f is defined as f(x) = 3 + lnx where x > 0

Another function, g is defined for all x as g(x) = ex4

Show that fg(x) = 4(1+ lnx)

So I've got up to fg(x) = e(3+ lnx)4, but no idea where to go from here to prove the above.

Would appreciate any support! :biggrin:
Original post by Funky_Giraffe
The function f is defined as f(x) = 3 + lnx where x > 0

Another function, g is defined for all x as g(x) = ex4

Show that fg(x) = 4(1+ lnx)

So I've got up to fg(x) = e(3+ lnx)4, but no idea where to go from here to prove the above.

Would appreciate any support! :biggrin:


That would be rather nasty to solve :tongue:

But what does fg(x) mean? Which function do you apply x to first?
Could you post an image of the actual question.
Original post by Funky_Giraffe
The function f is defined as f(x) = 3 + lnx where x > 0

Another function, g is defined for all x as g(x) = ex4

Show that fg(x) = 4(1+ lnx)

So I've got up to fg(x) = e(3+ lnx)4, but no idea where to go from here to prove the above.

Would appreciate any support! :biggrin:


remember fg(x) == f(g(x)) = 3 + ln(ex^4). apply the rule log(ab) = log(a) + log(b). also apply the rule log(a^2) = 2log(a) remember, if you apply a function to its inverse, everything cancels out.
Original post by Funky_Giraffe
The function f is defined as f(x) = 3 + lnx where x > 0

Another function, g is defined for all x as g(x) = ex4

Show that fg(x) = 4(1+ lnx)

So I've got up to fg(x) = e(3+ lnx)4, but no idea where to go from here to prove the above.

Would appreciate any support! :biggrin:


I may be mistaken, but I think you substituted f(x) for x into g(x) instead of g(x) for x into f(x) :tongue: From there, you can use log laws to get what they're looking for :smile:
If you meant ex^4 not e^x^4 then you can separate lnex^4 into lne+lnx^4
Original post by Kamara7
I may be mistaken, but I think you substituted f(x) for x into g(x) instead of g(x) for x into f(x) :tongue: From there, you can use log laws to get what they're looking for :smile:


That's it exactly! Thank you all for your help!
Original post by Kamara7
I may be mistaken, but I think you substituted f(x) for x into g(x) instead of g(x) for x into f(x) :tongue: From there, you can use log laws to get what they're looking for :smile:


Hang on, I still don't seem to be getting the same answer:

fg(x) = 3 + lnex4
= 3 + x4

which is not in the form that is being asked for.

Why can't I get this?!?!
Original post by Funky_Giraffe
Hang on, I still don't seem to be getting the same answer:

fg(x) = 3 + lnex4
= 3 + x4

which is not in the form that is being asked for.

Why can't I get this?!?!


It's actually less complicated than it looks:

[br]fg(x)=3+ln(ex4)[br]=3+ln(e)+ln(x4)[br]=3+1+4ln(x)[br]=4+4ln(x)[br]=4(1+ln(x))[br][br]fg(x) = 3+ln(ex^4)[br]= 3+ln(e)+ln(x^4)[br]= 3+1+4ln(x)[br]= 4+4ln(x)[br]= 4(1+ln(x))[br]

I think in this case it's easier to quickly show that solution to you. Makes sense now?
(edited 8 years ago)
Original post by sjgriffiths
It's actually less complicated than it looks:

[br]fg(x)=3+ln(ex4)[br]=3+ln(e)+ln(x4)[br]=3+1+4ln(x)[br]=4+4ln(x)[br]=4(1+ln(x))[br][br]fg(x) = 3+ln(ex^4)[br]= 3+ln(e)+ln(x^4)[br]= 3+1+4ln(x)[br]= 4+4ln(x)[br]= 4(1+ln(x))[br]

I think in this case it's easier to quickly show that solution to you. Makes sense now?


Yes!!! Thank you. Why do I never think to expand out the logarithm...

Quick Reply

Latest