- #1

jcsd

Science Advisor

Gold Member

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Is this correct (I was just a little worried that I've got a power of 1 that's not equal to 1)?:

1

proof:

using x = π in Euelr's identity:

e

e

square both sides:

e

putting both sides to the power of ln(5)/2πi

1

1

^{ln(5)/2πi}= 5proof:

using x = π in Euelr's identity:

e

^{xi}= cos(x) + isin(x)e

^{πi}= -1square both sides:

e

^{2πi}= 1putting both sides to the power of ln(5)/2πi

1

^{ln(5)/2πi}= (e^{2πi})^{ln(5)/2πi}= e^{ln(5)}= 5
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