How do you solve $e^{2 x} = 2 e^{x} + 1 = 0$ $e^(2x) = 2e^x + 1 = 0$ ?

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How do you solve $e^{2 x} = 2 e^{x} + 1 = 0$ $e^(2x) = 2e^x + 1 = 0$ ?

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Answer 1 · 2018-05-19T09:06:29

See explanation.

Explanation:

If you substitute the expression $e^{x}$ $e^x$ with a new variable you get a quadratic equation:

$e^{2 x} - 2 e^{x} + 1 = 0$ $e^(2x)-2e^x+1=0$

$t^{2} - 2 t + 1 = 0$ $t^2-2t+1=0$

${(t - 1)}^{2} = 0$ $(t-1)^2=0$

$t = 1$ $t=1$

Now we have to return to $x$ $x$ by solving:

$e^{x} = 1 \Rightarrow e^{x} = e^{0} \Rightarrow x = 0$ $e^x=1=>e^x=e^0=>x=0$

Answer: The equation has one solution $x = 0$ $x=0$ .

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How do you solve $e^{2 x} = 2 e^{x} + 1 = 0$ $e^(2x) = 2e^x + 1 = 0$ ?

1 Answer

Explanation:

$e^{2 x} - 2 e^{x} + 1 = 0$ $e^(2x)-2e^x+1=0$

$t^{2} - 2 t + 1 = 0$ $t^2-2t+1=0$

${(t - 1)}^{2} = 0$ $(t-1)^2=0$

$t = 1$ $t=1$

$e^{x} = 1 \Rightarrow e^{x} = e^{0} \Rightarrow x = 0$ $e^x=1=>e^x=e^0=>x=0$

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How do you solve e^(2x) = 2e^x + 1 = 0e2x=2ex+1=0e^(2x) = 2e^x + 1 = 0?

1 Answer

Explanation:

e^(2x)-2e^x+1=0e2x−2ex+1=0e^(2x)-2e^x+1=0

t^2-2t+1=0t2−2t+1=0t^2-2t+1=0

(t-1)^2=0(t−1)2=0(t-1)^2=0

t=1t=1t=1

e^x=1=>e^x=e^0=>x=0ex=1⇒ex=e0⇒x=0e^x=1=>e^x=e^0=>x=0

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Impact of this question

How do you solve $e^{2 x} = 2 e^{x} + 1 = 0$ $e^(2x) = 2e^x + 1 = 0$ ?

$e^{2 x} - 2 e^{x} + 1 = 0$ $e^(2x)-2e^x+1=0$

$t^{2} - 2 t + 1 = 0$ $t^2-2t+1=0$

${(t - 1)}^{2} = 0$ $(t-1)^2=0$

$t = 1$ $t=1$

$e^{x} = 1 \Rightarrow e^{x} = e^{0} \Rightarrow x = 0$ $e^x=1=>e^x=e^0=>x=0$