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Math Help - Help please ... Simple Exponential Function

  1. #1
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    Help please ... Simple Exponential Function

    hey guys i got my exam tomorrow and i cant nail this question

    my teacher wrote the answer = x=3/4 but that makes no sense to me ...

    thanks in advance
    Attached Thumbnails Attached Thumbnails Help please ... Simple Exponential Function-df.bmp  
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  2. #2
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    Quote Originally Posted by jimzer View Post
    hey guys i got my exam tomorrow and i cant nail this question

    my teacher wrote the answer = x=3/4 but that makes no sense to me ...

    thanks in advance
    Either 3^x = 0 (impossible for real values of x) or 3^x - 27 = 0 \Rightarrow 3^x = 27 = 3^3 \Rightarrow x = 3.
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    thnaks a lot!! is it because you can divide both sides by 3^x to get rid of the 3^x? and the 0 would still remain 0?
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    Quote Originally Posted by jimzer View Post
    thnaks a lot!! is it because you can divide both sides by 3^x to get rid of the 3^x? and the 0 would still remain 0?
    It's because if AB = 0 then one or both of A and B have to equal zero.
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  5. #5
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    Quote Originally Posted by jimzer View Post
    hey guys i got my exam tomorrow and i cant nail this question

    my teacher wrote the answer = x=3/4 but that makes no sense to me ...

    thanks in advance
    hey mate,

    for any equation of the form

    f(x)g(x) = 0

    either f(x) = 0, g(x) = 0 or f(x), g(x) = 0

    as f(x) = 3^x, it theoretically (and I am cringing as I write this) never can be 0 even as x approaches -infinity and as such for your situation the only solution can lie when g(x) = 3^x - 27 = 0
    thus,
    3^x = 27

    your teachers solution is incorrect as the only solution for this is x = 3 as 3^3 = 27 (this can easily be verified with a basic logarithmic analysis). Are you a high school student, if so what country are you residing in??? if Australian (as I am) never trust a 'maths' teachers word as gospel, the requirements to teach maths at a secondary level is pathetic.

    This should be your classroom


    Hope this helps,

    David
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    yeah true

    yeah dude im in hi school

    i live in noble park, go to st bedes in mentone

    and your right itd only work if the equation iwas 3^x(3^x) - 27 = 0 for x to equal 3/4

    thanks a lot
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    Quote Originally Posted by mr fantastic View Post
    It's because if AB = 0 then one or both of A and B have to equal zero.
    hey mate,

    just out of curiousity, if x wax complex, what values (and or set of solutions) would satisfy the equation??

    Cheers,

    David
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    Quote Originally Posted by David24 View Post
    hey mate,

    just out of curiousity, if x wax complex, what values (and or set of solutions) would satisfy the equation??

    Cheers,

    David
    None actually, because z = 0 is a singularity of the complex logarithm function.

    3^x = 0 is impossible for complex values of x too.
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    Hey Mr Fantastic,

    thats what I thought, just wanted to make sure, just out of interest, have you encountered a proof that any real number raised to a complex number can never be zero??? and moreover how can one evaluate a real number a raised to the complex value i ???

    Cheers,

    David
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  10. #10
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    Quote Originally Posted by David24 View Post
    Hey Mr Fantastic,

    thats what I thought, just wanted to make sure, just out of interest, have you encountered a proof that any real number raised to a complex number can never be zero??? and moreover how can one evaluate a real number a raised to the complex value i ???

    Cheers,

    David
    Like \left(e^{\pi}\right)^i \, ....
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    I'm assuming from your response that the direct way of addressing

    a^(i) is to simply use Euler's Identity

    e^(ix) = cos(x) + isin(x)

    thus as any real positive value a can be expressed as e^ln(a), i.e.
    a = e^(ln(a)

    then,
    a^(i) = (e^ln(a))^ i = e^(i * ln(a)) = cos(ln(a)) + i*sin(ln(a))

    Thus for,

    a^x = 0, where x can be complex, a is real and positive the expression becomes,
    let,
    x = c + di
    then
    a^x = 0 --> a^(c + di) = 0
    --> a^c * a^(di) = 0
    as c, d are real a^c ~= 0 thus we are left with
    a^(di) = 0 or (a^d)^i = 0
    as a, d are real, let a^d = f (nohing a^d is positive)
    hence we obtain,
    f^i = 0

    from before f^i = cos(ln(f)) + isin(ln(f)) = 0
    equating reals and imaginaries we have

    cos(ln(f)) = 0 , sin(ln(f)) = 0

    Thus, ln(f) = (2n+1)*pi/2 , ln(f) = n*pi , n being an integer
    So for a solution to exist to a^x = 0 then there must be an integer n such that,
    (2n+1)*pi/2 = n*pi,
    or,
    n*pi =pi/2 which has no real solutions for n being an integer.

    would you consider this to be a sufficient 'proof' and I say that word losely for why a^z = 0 is a is real, and if your in agreeance, what are you thoughts if a is complex???

    Regards,

    David
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