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Math Help - a^x=log_a_x

  1. #1
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    a^x=log_a_x

    At what value of a will the graph of A^x and log base(a) of (x) equal eachother?

    a^x=log_a_(x)?

    what i said was either one will intercept with the line y=x (Because since a^x and log_a_(x) are inverses. they will intersect at the line y=x

    so i made a^x=x, still couldnt solve

    the same with log_a_(x)=x
    Any ideas?

    ps. i use latex and winedt. is there a sticky on how to write that code through here or it won't work?

    cheers
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  2. #2
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    ps. i use latex and winedt. is there a sticky on how to write that code through here or it won't work?
    Yes, look at this thread: http://www.mathhelpforum.com/math-he...-tutorial.html .

    So instead of
    a^x=log_a_(x)
    you get
    a^x \;=\; \log_a x


    01


    EDIT: completely misread the question...
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  3. #3
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    Does the question place restrictions on a? Usually with logarithms, a>1, in which case the functions will never intercept...

    When you post, hit the button up on the toolbar that looks like a capital sigma. That will give you math tags. Type your LaTeX between these tags, and all will be well...
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  4. #4
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    Quote Originally Posted by sebko View Post
    At what value of a will the graph of A^x and log base(a) of (x) equal eachother?

    a^x=log_a_(x)?

    what i said was either one will intercept with the line y=x (Because since a^x and log_a_(x) are inverses. they will intersect at the line y=x

    so i made a^x=x, still couldnt solve

    the same with log_a_(x)=x
    Any ideas?

    ps. i use latex and winedt. is there a sticky on how to write that code through here or it won't work?

    cheers
    Consider the case that the graphs of f: f(x)=a^x and g: g(x)=\log_a(x) are tangent to each other.

    Then the tangent point must be placed on the line y=x and the gradient of both graphs must be 1 at the tangent point

    a^x=\log_a(x)~\wedge~a^x=x~\wedge~\ln(a)\cdot a^x=1

    Solve the 3rd equation for a^x and plug in this term into the first equation. Solve for a and x.

    I've got a = \sqrt[e]{e}
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  5. #5
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    Restrictions i believe were a>1 and a<=2
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  6. #6
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    Quote Originally Posted by sebko View Post
    Restrictions i believe were a>1 and a<=2
    My result fits miraculously into this interval:

    1 < \sqrt[e]{e} \approx 1.44466786... \leq  2
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  7. #7
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    Thankyou so much earboth... with your help i solved it myself aswell!

    I tried to put the LATEX that i did in here but i still wouldnt let me it said Syntax, even though i have latexd it with winedt , and saved it as a pdf with no errors... anyway. Here is my .tex file. and my .pdf


    .tex file
    problem.tex

    .pdf file
    problem.pdf
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