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Math Help - How to find a?

  1. #1
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    How to find a?

    How do you find b? In:

    -1/5 loga 3^-3 + 1/4 loga 3^1/4 = 53/320

    I have gotten as far as working out:

    loga 193/100 + loga 11/10 = 53/320 (at least I think I have done this part correctly)

    but I am unsure where to go from here. I think its something to do with y = loga(x) which is a^y = x but I'm unsure how to take it forward other than trial and error.
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  2. #2
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    Quote Originally Posted by JQ2009 View Post
    How do you find b? In:

    -1/5 loga 3^-3 + 1/4 loga 3^1/4 = 53/320

    I have gotten as far as working out:

    loga 193/100 + loga 11/10 = 53/320 (at least I think I have done this part correctly)

    but I am unsure where to go from here. I think its something to do with y = loga(x) which is a^y = x but I'm unsure how to take it forward other than trial and error.
    You should be able to write this as:

    log_a(3^{\frac{1}{8}})-log_a(3^{\frac{3}{5}})=\frac{53}{320}

    Then use the property log_a(x)-log_a(y)=log_a(\frac{x}{y}) to write this as a single logarithm.
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  3. #3
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    Should that not be:

    log(x1 * x2) = logx1 + logx2 as the negatives cancel each other out (or does the - not go to the power with the 3?) and 1/16 rather than 1/8? (appologies if I have missed something).

    Anyway doing it that way I got:

    1822/100 is that correct?
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  4. #4
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    Quote Originally Posted by JQ2009 View Post
    Should that not be:

    log(x1 * x2) = logx1 + logx2 as the negatives cancel each other out Yes you're right! (or does the - not go to the power with the 3?) and 1/16 rather than 1/8? (appologies if I have missed something).

    Anyway doing it that way I got:

    1822/100 is that correct?
    So this is what I end up with.

    log_a(3^{\frac{3}{5}+\frac{1}{16}})=\frac{53}{320}

    I don't see how you got 1822/100. I can't get a neat number like that. How did you proceed from what I wrote above?
    Last edited by adkinsjr; October 25th 2009 at 03:26 PM.
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  5. #5
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    I'm not sure now. I never wrote down my working for that part I punched it all into the calculator though I some how ended up with:

    loga 2122/1000 = 53/320

    so a^53/320 = 2122/1000

    so a = root 53/320 of 2122/1000 (I'm not sure if thats expressed correctly).

    But no matter what working I do now I cannot get 2122/1000, I must have made a mistake somewhere as doing what I though I'd done.

    Also I don't understand how you are getting 1/8 rather than a 1/16, could you explain that please?

    Anyway taking what you have up there I and doing what I did above I got 123.36
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  6. #6
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    Quote Originally Posted by JQ2009 View Post
    I'm not sure now. I never wrote down my working for that part I punched it all into the calculator though I some how ended up with:

    loga 2122/1000 = 53/320

    so a^53/320 = 2122/1000

    so a = root 53/320 of 2122/1000 (I'm not sure if thats expressed correctly).

    But no matter what working I do now I cannot get 2122/1000, I must have made a mistake somewhere as doing what I though I'd done.

    Also I don't understand how you are getting 1/8 rather than a 1/16, could you explain that please?

    Anyway taking what you have up there I and doing what I did above I got 123.36
    No, that was just another mistake on my part. I edited my last post.

    log_a(3^{\frac{3}{5}+\frac{1}{16}})=\frac{53}{320}

    If I add the fractions and pull out a calculator I get:

    log_a(3^{\frac{53}{80}})=log_a(3^{0.6625})

    So we can write:

    log_a(3^{0.6625})=\frac{53}{320}

    a^{\frac{53}{320}}=3^{0.6625}

    ln(a^{\frac{53}{320}})=ln(3^{0.6625})

    \frac{53}{320}ln(a)=0.6625ln(3)

    ln(a)=\frac{(320)(0.6625)ln(3)}{53}

    a=e^{\frac{(320)(0.6625)ln(3)}{53}}

    lol, that's the best I can do to isolate a. I can't tell if there is a more simplistic way to get a 'neat' answer.
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