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Math Help - A rather basic log question I believe

  1. #1
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    A rather basic log question I believe

    "Solve for x:"

    2 X 3^2x+3 = 3 X 2^3x+4

    (X is multiply)

    feel silly for posting this, is a simple log question i think, but its got me stumped.

    thanks
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  2. #2
    A Plied Mathematician
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    Just to clarify, you want to solve 2\cdot 3^{2x}+3=3\cdot 2^{3x}+4. Is that correct? If so, what steps have you taken?
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  3. #3
    Super Member bigwave's Avatar
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    it could be also:
    2(3)^{2x+3} = 3(2)^{3x+4}

    in which case x = -1
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  4. #4
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    sorry yeah the above is the way it's laid out - how did you get -1 ?
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  5. #5
    Super Member bigwave's Avatar
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    Cool well several ways to solve this..

    well several ways to solve this..

    but you can divide both sides by 2^1 and 3^1 to elimate the coeficients,

    then
    3^{2x+2} = 2^{3x+3} <br />
 \Rightarrow<br />
(2x+2)\ln{3} = (3x+3)\ln{2}<br />
\Rightarrow<br />
\frac{2x+2}{3x+3} = \frac{\ln{2}}{\ln{3}}<br />
    after calculation we are left with

    x(.107) = -.107\Rightarrow x= -1
    Last edited by bigwave; July 15th 2010 at 01:05 PM. Reason: added steps
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  6. #6
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    ahhh it's so simple, i'm so stupid. haha thanks, i just didn't start with the basic, "dividng by 2" part. oops.
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  7. #7
    Newbie Zharif93's Avatar
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    2(3)^2x+3 = 3(2)^3x+4

    log 2 log 3^(2x+3) = log 3 log 2^(3x+4)

    (2x+3) log 3 = log 3
    (3x+4) log 2 = log 2

    2x log 3 + 3 log = log 3

    2x log 3 = log 3 - 3 log 3

    x(2 log 3) = -2 log 3

    x = -2 log 3
    /2 log 3

    x = -1
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