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Math Help - indices problem

  1. #1
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    indices problem

    Solve the equation 3^{x+3}-3^{x+2}=2
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  2. #2
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    I seem to be having trouble working out this one... can someone show me what I did wrong?

    3^{x+3}-3^{x+2}=2

    \log{3^{x+3}}-\log{3^{x+2}}=\log{2}

    (x+3)\log{3}-(x+2)\log{3}=\log{2}

    x\log{3} + 3\log{3} - x\log{3} - 2\log{3} = \log{2}

    x\log{3}- x\log{3}  = \log{2}- 3\log{3} + 2\log{3}

    The x's cancel eachother out... and I don't know where went wrong.

    (By the way, x is a whole number)
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  3. #3
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    Hi, DavdM, log is NOT linear function : log(a+b) is NOT equal to log(a)+log(b).

    Here, we can write :

    3^{x+3}-3^{x+2}=2
    3^{x+2}.(3-1)=2
    3^{x+2}.2=2
    3^{x+2}=1
    finally, x=-2
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  4. #4
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    Quote Originally Posted by thorin View Post
    3^{x+2}.(3-1)=2
    I'm not getting where you got the (3-1) from. Please to explain?
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  5. #5
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    3^{x+3}-3^{x+2}=3.3^{x+2}-1.3^{x+2}=3^{x+2}.(3-1)
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  6. #6
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    Oh i get it.
    Last edited by DavidM; August 14th 2010 at 03:15 AM.
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  7. #7
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    Another way to do it: by the laws of exponents, 3^{x+ 3}= (3^x)(3^3)= 27(3^x) and 3^{x+ 2}= (3^x)(3^2)= 9(3^x). The equation becomes 3^{x+3}- 3^{x+ 2}= 27(3^x)- 9(3^x)= 18 (3^x)= 2 so that 3^x= \frac{2}{18}= \frac{1}{9}. Since 9= 3^2, \frac{1}{9}= 3^{-2} so we have, finally, 3^x= 3^{-2} and x= -2.
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