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Example: e to the two x is greater than

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e to the x plus twelve.

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Setting the right hand side equal to zero,

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we get e to the two x minus e to the x,

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minus twelve, is greater than zero.

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And then using our exponent rules,

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we know e to the two x is simply

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e to the x squared, minus e to the x, minus

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twelve is greater than zero. Now we have a

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quadratic, but instead of x's, we have

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e to the x. So factoring this quadratic,

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we get e to the x minus four, times

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e to the x plus three is greater than zero.

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So e to the x has to be equal to four or

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e to the x has to be equal to negative three,

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but we know e to the x cannot equal

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negative three, as it has to be greater

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than zero. So e to the x equal four is

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possible, by taking the natural log of

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both sides, we get x is equal to the natural

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log of four. Creating our sign chart,

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we only have the natural log of four.

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Looking at my two factors, we have

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e to the x minus four and e to the x plus three,

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which we know e to the x plus three is always

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positive because e to the x is always positive

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and we are adding three. And then

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e to the x minus four will be positive to

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the right, go through its zero, and then

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stay negative. So we have minus and plus

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and we are looking for only positive solutions,

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so we don't want to include our zero.

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So our answer is from the natural log of four,

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not included to infinity.