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Math Help - solving each equation using logs

  1. #1
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    solving each equation using logs

    i'm confused on:
    x=log (base 4)51.6

    and 5^2x=9^x-1



    thanks in advance.
    i think i'm just overthinking.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by beetz View Post
    i'm confused on:
    x=log (base 4)51.6
    there is nothing to solve here. you are simply stating some value for x. what do you want? and actual decimal expansion of a number?

    and 5^2x=9^x-1
    please clarify. for the 9, is the power x or is it (x - 1)?
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  3. #3
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    sorry it's 9^(x-1)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by beetz View Post
    5^2x=9^x-1
    5^{2x} = 9^{x - 1}

    take the log of both sides

    \Rightarrow \ln 5^{2x} = \ln 9^{x - 1}

    \Rightarrow 2x \ln 5 = (x - 1) \ln 9

    \Rightarrow 2x \ln 5 = x \ln 9 - \ln 9

    \Rightarrow 2x \ln 5 - x \ln 9 = - \ln 9

    \Rightarrow x (2 \ln 5 - \ln 9) = - \ln 9

    \Rightarrow x = \frac {- \ln 9}{2 \ln 5 - \ln 9}

    and you can simplify that a bit if you want
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  5. #5
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    Thanks so much!
    This problem is throwing me off:
    12^(x-4)=4^(2-x)
    so,
    (x-4)ln12=(2-x)ln4
    xln12 - 4ln12= 2ln4 - x ln 4
    ......now what?
    Thanks.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by beetz View Post
    Thanks so much!
    This problem is throwing me off:
    12^(x-4)=4^(2-x)
    so,
    (x-4)ln12=(2-x)ln4
    xln12 - 4ln12= 2ln4 - x ln 4
    ......now what?
    Thanks.
    don't let the logs confuse you. the log of a constant is a constant (provided the constant is greater then zero, of course). how would you do this if all the logs you saw were constants. you'd get the terms with x's on one side of the equation, and everything else on the other side. then you'd factor out the x and divide both sides by its coefficient
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  7. #7
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    ok:
    i added x ln4 to the left side, then
    2x ln16 - 4 ln12=2 ln4
    2x ln16=6 ln16
    2x=6
    x=3
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by beetz View Post
    ok:
    i added x ln4 to the left side, then
    2x ln16 - 4 ln12=2 ln4
    2x ln16=6 ln16
    2x=6
    x=3
    no. you cannot add the x's and the ln's like that. xln12 + xln4 is not the same as 2xln16

    keep them separate and factor out the x
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  9. #9
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    x ln12 - 4 ln12 = 2 ln4 - x ln4
    +x ln4 +x ln4
    x(ln12 + 1n4)- 4 ln12=2 ln4
    then divide both sides by ln12 + ln4 - 4 ln12?
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  10. #10
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    Quote Originally Posted by beetz View Post
    x ln12 - 4 ln12 = 2 ln4 - x ln4
    +x ln4 +x ln4
    x(ln12 + 1n4)- 4 ln12=2 ln4
    then divide both sides by ln12 + ln4 - 4 ln12?
    No. Keep only the x terms on the LHS.
    x(ln(12) + ln(4))- 4 ln(12)=2 ln(4)

    x(ln(12) + ln(4)) = 2 ln(4) + 4 ln(12)

    Now divide by what to get x by itself?

    -Dan
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  11. #11
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    Quote Originally Posted by beetz View Post
    i'm confused on:
    x=log (base 4)51.6
    thanks in advance.
    i think i'm just overthinking.
    This can be calculated as:

    \frac {\log(51.6)}{\log(4)} = 2.84464958
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