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Math Help - e^(2x)+4e^x-12=0 find the value of x that satisfies the equation

  1. #1
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    Talking e^(2x)+4e^x-12=0 find the value of x that satisfies the equation

    By drawing a graph I find x is about 0.693... but I need to show how to work it out using an algebraic method. I think I'm stuck on trying to get one x on one side of the equation if that makes sense. Please help! thanks

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  2. #2
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    As you can see, it is a quadratic. To make it clearer, let u = e^x, therefore:

    u^2+4u-12=0

    Now can you solve it?
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  3. #3
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    Hello, Alex!

    Solve: . e^{2x} + 4e^x - 12 \:=\:0

    Factor: . \left(e^x+6\right)\left(e^x - 2\right) \;=\;0

    Then we have: . \begin{Bmatrix}e^x+6 \:=\:0 & \Rightarrow & e^x \:=\:-6 & & \text{Impossible} \\ \\[-3mm]<br />
e^x - 2 \:=\:0 & \Rightarrow & e^x \:=\:2 & \Rightarrow & \boxed{x \:=\:\ln(2)}\;\; \end{Bmatrix}

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