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Math Help - [SOLVED] logrithimic question

  1. #1
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    [SOLVED] logrithimic question

    it's



    5^{x+1}=2(3^{2x})

    I have used base 10
    and Im at x=log_5-log_2 over 2log_3-log_5/2

    I expanded the bottom and cancelled -log_2

    so I have log_5 over 2log3-log_5

    I cant cancell the log base 5 because you cant make them have the same sign (neg. pos.) correct?

    please help

    thank you.
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  2. #2
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    Quote Originally Posted by brentwoodbc View Post
    it's



    5^{x+1}=2(3^{2x})

    I have to convert it to log form, then I think I have to use the formula
    log_a over log_b
    but I dont know how to convert something to logs with something on two sides of the = sign,

    please help

    thank you.
    Multiply out that two to give 6^{2x}

    Taking logs of both sides (I will choose base e) and using the rule

    k ln(x) = ln(x^k)

    (x+1)ln(5) = 2x ln(6) (using base 10 would give you (x+1)log_{10}5 = 2xlog_{10}6 and you can solve it in the same way)

    Can you solve it from there? I don't see why you'd use the change of base rule for this problem

    Spoiler:
    2xln(6) = xln(5)  + ln(5)

    2xln(6)-xln(5) = ln(5)

    x(2ln(6)-ln(5)) = ln(5)

    x = \frac{ln(5)}{2ln(6)-ln(5)}
    Last edited by e^(i*pi); April 7th 2009 at 12:13 PM. Reason: Completing Answer inside spoiler tags
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  3. #3
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    Quote Originally Posted by e^(i*pi) View Post
    Multiply out that two to give 6^{2x}

    Taking logs of both sides (I will choose base e) and using the rule

    k ln(x) = ln(x^k)

    (x+1)ln(5) = 2x ln(6) (using base 10 would give you (x+1)log_{10}5 = 2xlog_{10}6 and you can solve it in the same way)

    Can you solve it from there? I don't see why you'd use the change of base rule for this problem
    I thought you couldnt multiply something out if it has a differnt power?

    ie 2x2^2 = 8 not 16
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  4. #4
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    Quote Originally Posted by brentwoodbc View Post
    I thought you couldnt multiply something out if it has a differnt power?

    ie 2x2^2 = 8 not 16
    You can multiply the base but not the exponent. In the example you just put the base is 2 which would give 8. It is the same as 2^2 + 2^2 = 4 + 4 = 8.

    If I were to do 3(2^2) which is obviously 12 it would be 2^2+2^2+2^2 = 4+4+4 = 12.

    If it were a(b^n) then I'd get ab^n = b^n + b^n ... + b^n up to n=a
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    hm. my answer comes out close to yours. But the answer they give is

    x= log2-log5 over log5-2log3
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  6. #6
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    Quote Originally Posted by brentwoodbc View Post
    hm. my answer comes out close to yours. But the answer they give is

    x= log2-log5 over log5-2log3
    <br />
5^{x+1}=2(3^{2x})<br />

    taking logs of both sides

    (x+1)log(5) = log(2(3^2x))

    we can simplify the right hand side by using the law log{(ab)} = log(a) + log(b) with a = 2 and b = 3^2x. Also expanding the left

    xlog(5) + log(5) = log(2) + 2xlog(3)

    combine x terms:

    -2xlog(3) + xlog(5) = log(2)-log(5)

    Factor

    x(-2log(3) + log(5)) = log(2)-log(5)

    and so x = \frac{log(2)-log(5)}{log(5)-2log(3)}
    Last edited by e^(i*pi); April 7th 2009 at 12:32 PM. Reason: sign error
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  7. #7
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    Thank you, I appreciate it

    I did it a (slightly different way) and got

    log5-log2 over 2log3-log5

    is that the same thing? I know sometimes with fractions you can swap positions and negative signs and have the same value in a different order.

    your way makes sense though so thanks agian.
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  8. #8
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    Quote Originally Posted by brentwoodbc View Post
    Thank you, I appreciate it

    I did it a (slightly different way) and got

    log5-log2 over 2log3-log5

    is that the same thing? I know sometimes with fractions you can swap positions and negative signs and have the same value in a different order.

    your way makes sense though so thanks agian.
    Yeah, that's the same answer. It's the same as taking the negative of top and bottom which would cancel.
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