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Math Help - Please help, solving exponential equations

  1. #1
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    Unhappy Please help, solving exponential equations

    Please, show me how to do the steps for this equation, and the answer

    2 to the power x which also has an exponent which is 2 = 32(2 which has an exponent 4x)

    also, how do I put mathetical symbols into forums?
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  2. #2
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    Hello, needhelpplease1!

    Solve: . 2^{x^2} \;=\;32^{4x}

    We have: . 2^{x^2} \;=\;\left(2^5\right)^{4x}

    . . . . . . . . 2^{x^2} \;=\;2^{20x}

    . - Then: . x^2 \;=\;20x \quad\Rightarrow\quad x^2-20x\:=\:0 \quad\Rightarrow\quad x(x-20)\:=\:0


    Therefore: . x\;=\;0,\;20

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  3. #3
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    that's not the equation I meant :(

    I'm sorry, I know that it was hard to understand, but that's not the equation I meant. This is what I need help with:

     2^ {x^2} \;=\;32(2^{4x})
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  4. #4
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    sorry, but now the equation I meant!

    Quote Originally Posted by Soroban View Post
    Hello, needhelpplease1!


    We have: . 2^{x^2} \;=\;\left(2^5\right)^{4x}

    . . . . . . . . 2^{x^2} \;=\;2^{20x}

    . - Then: . x^2 \;=\;20x \quad\Rightarrow\quad x^2-20x\:=\:0 \quad\Rightarrow\quad x(x-20)\:=\:0


    Therefore: . x\;=\;0,\;20

    SO SO sorry, I know it was confusing to read but what I meant was:

     2^ {x^2} \;=\;32(2^{4x})
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