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Math Help - Solve the equation...

  1. #1
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    Solve the equation...

    3^(x-2) = 4^(2-x)

    I know how to do these when the bottom numbers can be divided by each other to get the top equations down to solve, but how can I do this when when the numbers on the bottom are different?

    I was thinking to times the 3 and 4 to get,

    12^(4x-8) = 12^(6-3x)

    and then I could do it from there. But am I even going about it the way I have to?

    thanks
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by driverfan2008 View Post
    3^(x-2) = 4^(2-x)

    I know how to do these when the bottom numbers can be divided by each other to get the top equations down to solve, but how can I do this when when the numbers on the bottom are different?

    I was thinking to times the 3 and 4 to get,

    12^(4x-8) = 12^(6-3x)

    and then I could do it from there. But am I even going about it the way I have to?

    thanks
    Do you know how to apply logarithms?

    3^{x-2}=4^{2-x}\implies (x-2)\log(3)=(2-x)\log(4).

    Can you try taking it from here?

    --Chris
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  3. #3
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    Hello, driverfan2008!

    3^{x-2} \:= \:4^{2-x}

    With two different bases, I was going to suggest using logs . . .
    . . then I saw something!


    We have: . 3^{x-2} \;=\;4^{2-x}

    . . . . . . . . 3^{x-2} \;=\;4^{-(2-x)}

    . . . . . . . . 3^{x-2} \;=\;\frac{1}{4^{x-2}}

    . . . (3^{x-2})(4^{x-2}) \;=\;1

    . . . . . . . 12^{x-2} \;=\;1

    . . . . . . . 12^{x-2} \;=\;12^0

    . . . . . . . x - 2 \;=\;0

    . . . . . . . . . . \boxed{x \;=\;2}

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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, driverfan2008!


    With two different bases, I was going to suggest using logs . . .
    . . then I saw something!


    We have: . 3^{x-2} \;=\;4^{2-x}

    . . . . . . . . 3^{x-2} \;=\;4^{-(2-x)}

    . . . . . . . . 3^{x-2} \;=\;\frac{1}{4^{x-2}}

    . . . (3^{x-2})(4^{x-2}) \;=\;1

    . . . . . . . 12^{x-2} \;=\;1

    . . . . . . . 12^{x-2} \;=\;12^0

    . . . . . . . x - 2 \;=\;0

    . . . . . . . . . . \boxed{x \;=\;2}

    I was thinking about that...
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