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

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

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Answer 1 · 2018-05-19T09:06:29

See explanation.

Explanation:

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

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

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

$(t-1)^2=0$

$t=1$

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

$e^x=1=>e^x=e^0=>x=0$

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

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

1 Answer

Explanation:

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

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

$(t-1)^2=0$

$t=1$

$e^x=1=>e^x=e^0=>x=0$

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

1 Answer

Explanation:

e^(2x)-2e^x+1=0e^(2x)-2e^x+1=0

t^2-2t+1=0t^2-2t+1=0

(t-1)^2=0(t-1)^2=0

t=1t=1

e^x=1=>e^x=e^0=>x=0e^x=1=>e^x=e^0=>x=0

Related questions

Impact of this question

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

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

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

$(t-1)^2=0$

$t=1$

$e^x=1=>e^x=e^0=>x=0$