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Algebra and Functions Help

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Original post by emx_eco
ooooooh therefore the domain is x > 3. I get it now thank you!!

Do you think you could help me with part b) -


f(x) = e^(2x) + 3 where x is any real number
g(x) = ln(x-1) where x is any real number and x > 1

Find fg(x) and state its range


Have you made a start?
Reply 21
Original post by _gcx
Have you made a start?


so at the moment I have:

fg(x) = eln(x-1)^2 + 3
Original post by emx_eco
so at the moment I have:

fg(x) = eln(x-1)^2 + 3


Not quite.

You should have have fg(x)=e2ln(x1)+3fg(x) = e^{2\ln(x-1)} + 3. You can write this as (eln(x1))2+3\left(e^{\ln(x-1)}\right)^2 + 3. Do you know what to do from there?

(unless you meant ln[(x1)2]\ln[(x-1)^2] or (eln(x1))2+3\left(e^{\ln(x-1)}\right)^2 + 3, then you're fine)
(edited 6 years ago)
Reply 23
Original post by _gcx
Not quite.

You should have have fg(x)=e2ln(x1)+3fg(x) = e^{2\ln(x-1)} + 3. You can write this as (eln(x1))2+3\left(e^{\ln(x-1)}\right)^2 + 3. Do you know what to do from there?

(unless you meant ln[(x1)2]\ln[(x-1)^2], then you're fine)


how can you write it in that format? what is the rule?
Original post by emx_eco
how can you write it in that format? what is the rule?


abc(ab)ca^{bc} \equiv (a^b)^c. You also have bln(a)ln(ab)b\ln(a) \equiv \ln(a^b).
Reply 25
Original post by _gcx
abc(ab)ca^{bc} \equiv (a^b)^c. You also have bln(a)ln(ab)b\ln(a) \equiv \ln(a^b).


oh okay I understand that now. So whats comes next? :smile:
Original post by emx_eco
oh okay I understand that now. So whats comes next? :smile:


You should be fine that e2ln(x1)+3=(x1)2+3e^{2\ln(x-1)} + 3 = (x-1)^2 + 3. Are you familiar with how to proceed?
Reply 27
Original post by _gcx
You should be fine that e2ln(x1)+3=(x1)2+3e^{2\ln(x-1)} + 3 = (x-1)^2 + 3. Are you familiar with how to proceed?


how do these two things equal each other?
Original post by emx_eco
how do these two things equal each other?


eln(x1)=x1e^{\ln(x-1)} = x-1 :smile:
Reply 29
Original post by _gcx
eln(x1)=x1e^{\ln(x-1)} = x-1 :smile:


how come the e cancels out? (sorry for so many questions!)
Original post by _gcx
eln(x1)=x1e^{\ln(x-1)} = x-1 :smile:


Does it??
Original post by Ano9901whichone
Does it??


yes... it is given in the question that x>1x>1. Please don't confuse the OP, even if your comment was in jest.
(edited 6 years ago)
Original post by emx_eco
how come the e cancels out? (sorry for so many questions!)


because ln\ln is the inverse of exe^x :smile: A property of inverse functions is that f(f1(x))=f1(f(x))=xf(f^{-1}(x)) = f^{-1}(f(x)) = x.
Reply 33
Original post by _gcx
because ln\ln is the inverse of exe^x :smile: A property of inverse functions is that f(f1(x))=f1(f(x))=xf(f^{-1}(x)) = f^{-1}(f(x)) = x.


ohhhhh! So now that you have (x-1)^2 +3 you just sketch it with the minimum point (1,3) meaning the range is f(x) > 1? (thats meant to be the bigger than or equal to sign with the line underneath)
Original post by emx_eco
ohhhhh! So now that you have (x-1)^2 +3 you just sketch it with the minimum point (1,3) meaning the range is f(x) > 1? (thats meant to be the bigger than or equal to sign with the line underneath)


Close, that's the domain. (x>1x>1, remember that ln0\ln 0 is undefined) The range is fg(x)>3fg(x) > 3. Can you see why?
Reply 35
Original post by _gcx
Close, that's the domain. (x>1x>1, remember that ln0\ln 0 is undefined) The range is fg(x)>3fg(x) > 3. Can you see why?


oh wait yh, my mistake! Thanks so much for all your help :smile: xxx

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