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    1st off i want to start with log....

    i understand:
    if e^x = 3........... Then X = ln 3

    But when its:

    (2e^ 4x) - 3 = 8

    What happens to the 2 in front of the e.

    I thought: 2 (4x) - ln3 = ln 8 .... but don't think thats right.

    2nd.

    How to find the inverse. I understand for simple function...but when its
    F(x) = (2e^2x+4), for all real numbers.

    3. domain is the measure of the x value.

    and 4. do you have to learn all the trig identities or are some given in the formula sheet.


    Thank you very much.Kam.
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    (Original post by kam_007)
    1st off i want to start with log....

    i understand:
    if e^x = 3........... Then X = ln 3

    But when its:

    (2e^ 4x) - 3 = 8

    What happens to the 2 in front of the e.

    I thought: 2 (4x) - ln3 = ln 8 .... but don't think thats right.
    2nd. Is that suppose to be 2e^{2x}+4 or 2e^{2x+4}

    3rd. Are you trying to find the domain?

    4. It's a good idea to learn them
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    ok... here's the solution

    (2e^4x)-3= 8 must be treated like any other algebraic problem as in you have to isolate the term containing x first.

    So, first you add 3 on both sides so as to eliminate it from the left hand side (LHS)

    now you have 2e^4x=11

    divide both sides by 2 and you get e^4x= 11/2

    now, ln both sides and you get 4x= ln (11/2)
    therefore x = 1/4 ln (11/2) which can also be written as x= ln(11/2)^(1/4)

    hope that cleared your doubt (or did i confuse you more??? :P)
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    (Original post by whizkid)
    ok... here's the solution

    (2e^4x)-3= 8 must be treated like any other algebraic problem as in you have to isolate the term containing x first.

    So, first you add 3 on both sides so as to eliminate it from the left hand side (LHS)

    now you have 2e^4x=11

    divide both sides by 2 and you get e^4x= 11/2

    now, ln both sides and you get 4x= ln (11/2)
    therefore x = 1/4 ln (11/2) which can also be written as x= ln(11/2)^(1/4)

    hope that cleared your doubt (or did i confuse you more??? :P)
    Duhhhhhhhhhhhhhhh

    I feel like such a pillock!!

    You just get rid of the front term.......Thanks alot damn im dumb.
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    ANy thoughts about the inverse function one...??!
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    Lol... i've had a whole year's worth of practice. You are not dumb! I found it pretty confusing when i started as well!!
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    2. Long time since I done C3 now

    y = 2e^{2x+4} \\



\dfrac{y}{2} = e^{2x+4} \\



2x + 4 = ln[\dfrac{y}{2}] ...

    Spoiler:
    Show
    2x = ln[\dfrac{y}{2}] - 4 \\



x = \dfrac{ln[\dfrac{y}{2}]}{2} - 2


    So try to get x on its own and then that is your inverse function after you replace the x with f'(x) and you replace all y's with x's.

    Answer
    \therefore F'(x) = \dfrac{ln[\dfrac{x}{2}]}{2} - 2
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    Thanks a bunch got it.... but how do you work out the domain. isit the range of the orginal function?
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    ok... for inverses, you know that you have to replace x with y and vice versa right??

    So, let F(X) be y
    therefore, your problem is now, y= (2e^2x+4)
    now, you replace x with y and y with x

    so, now you have, x= (2e^2y+4)

    1/2 x = (e^2y+4)

    ln (1/2 x) =2y+4
    so, 2y = ln(1/2 x)-4
    y= 1/2 ln (1/2 x)-2

    so, y= (ln (1/2 x)^1/2 ) -2
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    (Original post by kam_007)
    Thanks a bunch got it.... but how do you work out the domain. isit the range of the orginal function?
    yes usually
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    Thanks got it......

    Got a fractions question:

    (x + (1/x) - 2) / (x - 1)......simplify??

    answer is (x-1) / x
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    \dfrac{x + \dfrac{1}{x} - 2}{x - 1}

    I'll edit this but its just clearer for me to see.

    Ok:

    time the whole thing by x/x:

    \dfrac{x^2 + 1 - 2x}{x^2 - x}

    Now it depends which way you want to do this. You can do algebraic long division or you can do that A, B, C method (forgotten the name).
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    \frac{x+\frac{1}{x}-2}{x-1}=\frac{x^2-2x+1}{x-1}=\frac{(x-1)(x-1)}{x-1}=x-1

    I believe the answer is x-1
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    (Original post by boy)
    \frac{x+\frac{1}{x}-2}{x-1}=\frac{x^2-2x+1}{x-1}=\frac{(x-1)(x-1)}{x-1}=x-1

    I believe the answer is x-1
    You've multiplied the top by x but not the bottom

    \frac{x+\frac{1}{x}-2}{x-1}=\frac{x^2-2x+1}{x(x-1)}=\frac{(x-1)(x-1)}{x(x-1)}=\frac{x-1}{x} = 1 - \frac{1}{x}

    If something factorises, factorise it
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    (Original post by boy)
    \frac{x+\frac{1}{x}-2}{x-1}=\frac{x^2-2x+1}{x-1}=\frac{(x-1)(x-1)}{x-1}=x-1

    I believe the answer is x-1
    I get it.....

    but you would have to times top and bottom by x...which would give you x-1 / x

    x^2 - 2x + 1 / x^2 - x

    (x - 1) (x - 1) / x (x-1)

    x - 1 / x
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    (Original post by sohanshah)
    \dfrac{x + \dfrac{1}{x} - 2}{x - 1}

    I'll edit this but its just clearer for me to see.

    Ok:

    time the whole thing by x/x:

    \dfrac{x^2 + 1 - 2x}{x^2 - x}

    Now it depends which way you want to do this. You can do algebraic long division or you can do that A, B, C method (forgotten the name).
    How'd you guys get the umber cool..... real letters, to make it so clear.
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    Sorry, my mistake
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    You mean Latex.

    Put everything in [ latex ] tags

    http://www.thestudentroom.co.uk/wiki/LaTex
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    (Original post by kam_007)
    How'd you guys get the umber cool..... real letters, to make it so clear.
    It's called latex.

    Use [latex][/latex] tags and put the maths in the middle.

    Here's the guide to it.
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    Can anyone explain range and domain to me. Its now the one thing im getting stuck on......

    Also do you have to learn all the trigonometric identities?
 
 
 
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