How do solve $3 e^{x} = 2 e - x + 4$ $3e^x=2e-x+4$ ?

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How do solve $3 e^{x} = 2 e - x + 4$ $3e^x=2e-x+4$ ?

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Answer 1 · 2015-11-08T21:09:44

If you mean $3 e^{x} = 2 e^{- x} + 4$ $3e^x=2e^(-x)+4$ , then rearrange as a quadratic in $e^{x}$ $e^x$ , solve and take logs to find:

$x = ln (\frac{2 + \sqrt{10}}{3})$ $x = ln((2+sqrt(10))/3)$

Explanation:

Assuming you meant $3 e^{x} = 2 e^{- x} + 4$ $3e^x=2e^(-x)+4$ , first multiply by $e^{x}$ $e^x$ to get:

$3 {(e^{x})}^{2} = 2 + 4 (e^{x})$ $3(e^x)^2 = 2+4(e^x)$

Subtract the right hand side from the left to get:

$3 {(e^{x})}^{2} - 4 (e^{x}) - 2 = 0$ $3(e^x)^2-4(e^x)-2 = 0$

Using the quadratic formula, we get:

$e^{x} = \frac{4 \pm \sqrt{4^{2} - (4 \times 3 \times - 2)}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 + 24}}{6}$ $e^x = (4+-sqrt(4^2-(4xx3xx-2)))/(2*3) =(4+-sqrt(16+24))/6$

$= \frac{4 + = \sqrt{40}}{6} = \frac{4 \pm 2 \sqrt{10}}{6} = \frac{2 \pm \sqrt{10}}{3}$ $=(4+=sqrt(40))/6 = (4+-2sqrt(10))/6 = (2+-sqrt(10))/3$

Now $\sqrt{10} > 2$ $sqrt(10) > 2$ , so to get a positive value for $e^{x}$ $e^x$ (and hence a Real value for $x$ $x$ ) we need to choose the $+$ $+$ sign here to find:

$e^{x} = \frac{2 + \sqrt{10}}{3}$ $e^x = (2+sqrt(10))/3$

and hence:

$x = ln (\frac{2 + \sqrt{10}}{3})$ $x = ln((2+sqrt(10))/3)$

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How do solve $3 e^{x} = 2 e - x + 4$ $3e^x=2e-x+4$ ?

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