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Math Help - Log Help

  1. #1
    Junior Member
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    Log Help

    Seems to be a lot of log help going on tonight.. here's one that im not sure where to start..



    also,


    Now on this on i figured out that i need to convert to exponential form this gives me:

    4^4=x^2/(x+5)

    then i multiply out:

    256x+1280=x^2

    rearrange:
    x^2-256x-1280=0

    now i figure punch it into the quadratic formula and call it a day? the program i'm putting it into for HW says it's only 1/2 right if i put in (256+sqrt(70656))/2 and yes i tried  (256-sqrt(70656))/2 since it is +/-... ideas?

    oh and by the way, how do i put square root symbols in these messages?
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  2. #2
    MHF Contributor
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    Quote Originally Posted by xsavethesporksx View Post
    Hint:

    Let k = log(5) / log(6)

    (6x - 3) / (4x - 6) = k
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  3. #3
    MHF Contributor

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    [QUOTE=xsavethesporksx;475365]Seems to be a lot of log help going on tonight.. here's one that im not sure where to start..



    Take the logarithm (either "common" or "natural"- whichever is simpler on your calculator) of both sides to get (6x-3)log(6)= (4x- 6)log(5) so that 6log(6)x- 3log(6)= 4log(5)x- 6log(5). Solve that linear equation for x. That is the same as saying \frac{6x- 3}{4x- 6}= \frac{log(5)}{log(6)} as Wilmer gives.

    also,


    Now on this on i figured out that i need to convert to exponential form this gives me:

    4^4=x^2/(x+5)

    then i multiply out:

    256x+1280=x^2

    rearrange:
    x^2-256x-1280=0

    now i figure punch it into the quadratic formula and call it a day? the program i'm putting it into for HW says it's only 1/2 right if i put in (256+sqrt(70656))/2 and yes i tried  (256-sqrt(70656))/2 since it is +/-... ideas?
    The quadratic equation gives two roots- and since squaring or adding 5 to the negative root gives a positive value in the square root, bothare valid solutions.

    oh and by the way, how do i put square root symbols in these messages?
    The best thing to do is to learn "LaTex". For example, [ math ]\int_{-\infty}^\infty e^{-x^2}dx= \sqrt{2\pi} [ /math ] without the spaces inside [ ] gives \int_{-\infty}^\infty e^{-x^2}dx= \sqrt{2\pi} .

    Clicking on any such formula will show you the code to produce it. There is a tutorial here:
    http://www.mathhelpforum.com/math-help/latex-help/
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