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Math Help - [SOLVED] Exponential Functions

  1. #1
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    Question [SOLVED] Exponential Functions

    I need help with 2 problems, please. I'm not sure how to solve them.

    #1:
    <br />
4x^{2}(2^{x}) - 9(2^{x}) = 0<br />

    #2:
    <br />
\frac{(x+4)10^{x}}{x-3} = 2x(10^{x})<br />

    Thanks for the help! I haven't done this kind of math in awhile, and I'm just confused.
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  2. #2
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    Quote Originally Posted by juicysharpie View Post
    I need help with 2 problems, please. I'm not sure how to solve them.

    #1:
    <br />
4x^{2}(2^{x}) - 9(2^{x}) = 0<br />

    #2:
    <br />
\frac{(x+4)10^{x}}{x-3} = 2x(10^{x})<br />

    Thanks for the help! I haven't done this kind of math in awhile, and I'm just confused.
    #1 Factorise: 2^x (4x^2 - 9) = 0. Now try solving for x.


    #2 Re-arrange: \frac{(x+4)10^{x}}{x-3} - 2x(10^{x}) = 0.

    Now factorise: 10^x \left( \frac{x + 4}{x - 3} - 2x \right) = 0

    \Rightarrow 10^x \left( \frac{x + 4}{x - 3} - \frac{2x (x - 3)}{x - 3} \right) = 0

    \Rightarrow 10^x \left( \frac{x + 4 - 2x(x - 3)}{x - 3}\right) = 0

    \Rightarrow 10^x \left( \frac{-2x^2 + 7x + 4}{x - 3}\right) = 0

    \Rightarrow 10^x \left( \frac{(2x + 4)(4 - x)}{x - 3}\right) = 0

    Now try solving for x.
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  3. #3
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    Thanks for the help!

    I just wanted to check my answers if that's O.K.

    #1: x = \frac{3}{2}

    and

    #2 x = -2 or -4

    Are those correct?
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  4. #4
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    Quote Originally Posted by juicysharpie View Post
    Thanks for the help!

    I just wanted to check my answers if that's O.K.

    #1: x = \frac{3}{2}

    and

    #2 x = -2 or -4

    Are those correct?
    Not quite.

    #1: 4x^2 - 9 = 0 \Rightarrow (2x - 3)(2x + 3) = 0 \Rightarrow x = \pm \frac{3}{2}. Note: 2^x = 0 has no real solution.


    #2: 2x + 4 = 0 \Rightarrow x = -2 or 4 - x = 0 \Rightarrow x = 4 (NOT -4).
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