2

this needs to be converted to exponential form and I can't seem to figure it out.

Any help is appreciated, thank you!

$$10 \log(1+i) = \log 2$$

Aditya Hase
  • 8,851
Ryan
  • 23

3 Answers3

1

Apply the basic logarithm property: $\log (1+i)^{10}=\log 2$, and $(1+i)^{10}=2$.

ajotatxe
  • 65,084
1

Note that:

  • $10\log(1+i)=\log((1+i)^{10})$

Therefore:

  • $10\log(1+i)=\log2\implies$

  • $\log((1+i)^{10})=\log2\implies$

  • $(1+i)^{10}=2\implies$

  • $1+i=\sqrt[10]{2}\implies$

  • ${i}=\sqrt[10]{2}-1$

barak manos
  • 43,109
0

Assuming that $$ \log x=\log_e x=\ln x $$ We have $$10 \log(1+i) = \log 2$$ $$ \log(1+i)^{10} = \log 2$$ $$ e^{\log(1+i)^{10}}= e^{\log 2}$$ $$ (1+i)^{10}= 2$$ $$ 1+i=\sqrt[10]{2} $$ $$ i=\sqrt[10]{2} -1$$ Where $i$ does not denote the imaginary unit and is just a poorly chosen variable name. If $i$ is intended to denote the imaginary unit, then this equation is false.

k170
  • 9,045