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Question Number 191203 by otchereabdullai last updated on 20/Apr/23

If 3^x =−1 find the value of x

$$\:{If}\:\:\mathrm{3}^{{x}} =−\mathrm{1}\:{find}\:{the}\:{value}\:{of}\:{x} \\ $$

Answered by Frix last updated on 20/Apr/23

3^x =−1 e^(xln 3) =e^(i(2n+1)π) xln 3 =i(2n+1)π x=i(((2n+1)π)/(ln 3))

$$\mathrm{3}^{{x}} =−\mathrm{1} \\ $$$$\mathrm{e}^{{x}\mathrm{ln}\:\mathrm{3}} =\mathrm{e}^{\mathrm{i}\left(\mathrm{2}{n}+\mathrm{1}\right)\pi} \\ $$$${x}\mathrm{ln}\:\mathrm{3}\:=\mathrm{i}\left(\mathrm{2}{n}+\mathrm{1}\right)\pi \\ $$$${x}=\mathrm{i}\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}{\mathrm{ln}\:\mathrm{3}} \\ $$

Answered by Skabetix last updated on 20/Apr/23

3^x =i^2 xln(3)=ln(i^2 ) x=((ln(i^2 ))/(ln(3)))

$$\mathrm{3}^{{x}} ={i}^{\mathrm{2}} \\ $$$${xln}\left(\mathrm{3}\right)={ln}\left({i}^{\mathrm{2}} \right) \\ $$$${x}=\frac{{ln}\left({i}^{\mathrm{2}} \right)}{{ln}\left(\mathrm{3}\right)} \\ $$

Commented by Skabetix last updated on 21/Apr/23

■ ln(i)=ln((−1)^(1/2) )=(1/2)ln(−1)=(1/2)×πi so ln(i^2 )=2×(1/2)×πi=πi

$$\blacksquare\:{ln}\left({i}\right)={ln}\left(\left(−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(−\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}×\pi{i} \\ $$$${so}\:{ln}\left({i}^{\mathrm{2}} \right)=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}×\pi{i}=\pi{i} \\ $$

Commented by Frix last updated on 21/Apr/23

The value of ln (i^2 ) =?

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{ln}\:\left(\mathrm{i}^{\mathrm{2}} \right)\:=? \\ $$

Answered by anr0h3 last updated on 21/Apr/23

−1=e^(i∙π) 3^x =e^(i∙π) ln(3^x )=ln(e^(i∙π) ) x∙ln(3)=i∙π x=((((√(−1)))∙(π))/(ln(3)))

$$ \\ $$$$−\mathrm{1}={e}^{{i}\centerdot\pi} \\ $$$$\mathrm{3}^{{x}} ={e}^{{i}\centerdot\pi} \\ $$$${ln}\left(\mathrm{3}^{{x}} \right)={ln}\left({e}^{{i}\centerdot\pi} \right) \\ $$$${x}\centerdot{ln}\left(\mathrm{3}\right)={i}\centerdot\pi \\ $$$${x}=\frac{\left(\sqrt{−\mathrm{1}}\right)\centerdot\left(\pi\right)}{{ln}\left(\mathrm{3}\right)} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Frix last updated on 21/Apr/23

−1=e^((2n+1)πi) ln (−1) =(2n+1)πi Complex logaritms are not unique [why do you use (√(−1)) instead of i at the end?]

$$−\mathrm{1}=\mathrm{e}^{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi\mathrm{i}} \\ $$$$\mathrm{ln}\:\left(−\mathrm{1}\right)\:=\left(\mathrm{2}{n}+\mathrm{1}\right)\pi\mathrm{i} \\ $$$$\mathrm{Complex}\:\mathrm{logaritms}\:\mathrm{are}\:\mathrm{not}\:\mathrm{unique} \\ $$$$\left[\mathrm{why}\:\mathrm{do}\:\mathrm{you}\:\mathrm{use}\:\sqrt{−\mathrm{1}}\:\mathrm{instead}\:\mathrm{of}\:\mathrm{i}\:\mathrm{at}\:\mathrm{the}\:\mathrm{end}?\right] \\ $$

Commented by anr0h3 last updated on 21/Apr/23

that's correct, ln(-1) = ln(i^2) and ln(i^2)=(2n+1)πi

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