Imaginary Solution

Algebra Level 3

Which of the following is not a way to represent the value of "i"

ln(e^(ln(-1)))/pi ln(-1)/pi sqrt(-4)/2 sqrt(-1) sqrt(e^ln(-1)) ln(e^(sqrt(-1)))/pi

Joshua Crawford by Joshua Crawford , 15, Australia

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1 solution

Jordan Cahn
Jan 29, 2019

ln ⁡ e − 1 π = − 1 π = i π ≠ i \dfrac{\ln e^{\sqrt{-1}}}{\pi} = \dfrac{\sqrt{-1}}{\pi}=\dfrac{i}{\pi}\neq i

On the other hand,

  • − 1 = i \sqrt{-1} = i
  • − 4 2 = 2 i 2 = i \dfrac{\sqrt{-4}}{2} = \dfrac{2i}{2} = i
  • e ln ⁡ ( − 1 ) = − 1 = i \sqrt{e^{\ln (-1)}} = \sqrt{-1} = i
  • ln ⁡ ( − 1 ) π = i π π = i \dfrac{\ln(-1)}{\pi} = \dfrac{i\pi}{\pi} = i , since e i π = − 1 e^{i\pi}=-1
  • ln ⁡ e ln ⁡ ( − 1 ) π = ln ⁡ ( − 1 ) π = i π π = i \dfrac{\ln e^{\ln (-1)}}{\pi} = \dfrac{\ln(-1)}{\pi} = \dfrac{i\pi}{\pi} = i
0 Helpful 0 Interesting 0 Brilliant 0 Confused

That's everything :) (except you put Pi at the end of the last equation instead of i)

Joshua Crawford - 2 years, 4 months ago

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Good catch. Fixed!

Jordan Cahn - 2 years, 4 months ago

well since you said mine was wrong all of your problems will now be wrong to me tooo. You have made my Name dishonOred in the BrilIIant Community.

EswaR Gogineni - 2 years, 4 months ago

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? what did I say was wrong. lol

Joshua Crawford - 2 years, 4 months ago

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