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Question Number 204512 by Ghisom last updated on 20/Feb/24

solve for x∈C 3^(2ix) −3^(ix) 2+5=0

$$\mathrm{solve}\:\mathrm{for}\:{x}\in\mathbb{C} \\ $$$$\mathrm{3}^{\mathrm{2i}{x}} −\mathrm{3}^{\mathrm{i}{x}} \mathrm{2}+\mathrm{5}=\mathrm{0} \\ $$

Answered by Rasheed.Sindhi last updated on 20/Feb/24

(3^(ix) )^2 −2(3^(ix) )+1+4=0 (3^(ix) −1)^2 =−4 3^(ix) −1=±2i 3^(ix) =1±2i ln(3^(ix) )=ln(1±2i) x=((ln(1±2i) )/(iln 3 )) x=−((iln(1±2i) )/(ln 3 ))

$$\left(\mathrm{3}^{{ix}} \right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{3}^{{ix}} \right)+\mathrm{1}+\mathrm{4}=\mathrm{0} \\ $$$$\left(\mathrm{3}^{{ix}} −\mathrm{1}\right)^{\mathrm{2}} =−\mathrm{4} \\ $$$$\mathrm{3}^{{ix}} −\mathrm{1}=\pm\mathrm{2}{i} \\ $$$$\mathrm{3}^{{ix}} =\mathrm{1}\pm\mathrm{2}{i} \\ $$$$\mathrm{ln}\left(\mathrm{3}^{{ix}} \right)=\mathrm{ln}\left(\mathrm{1}\pm\mathrm{2}{i}\right)\: \\ $$$${x}=\frac{\mathrm{ln}\left(\mathrm{1}\pm\mathrm{2}{i}\right)\:}{{i}\mathrm{ln}\:\mathrm{3}\:}\: \\ $$$${x}=−\frac{{i}\mathrm{ln}\left(\mathrm{1}\pm\mathrm{2}{i}\right)\:}{\mathrm{ln}\:\mathrm{3}\:}\: \\ $$

Commented by Ghisom last updated on 20/Feb/24

thank you

Answered by Frix last updated on 20/Feb/24

With x=a+bi; a, b ∈R we get: Real part: (1/9^b )(2cos^2 (aln 3) −3^b 2cos (aln 3) +9^b 5−1)=0 Imaginary part: (2/9^b )(cos (aln 3) −3^b )sin (aln 3) =0 (1/9^b )≠0 ⇒ { (((1) cos^2 (aln 3) −3^b cos (aln 3) +((9^b 5−1)/2)=0)),(((2) (cos (aln 3) −3^b )sin (aln 3) =0)) :} (2) With sin (aln 3) =0 ⇒ b∉R cos (aln 3) =3^b ⇒ (1) 9^b =(1/5) ⇒ b=−((ln 5)/(2ln 3)) ⇒ cos (aln 3) =((√5)/5) ⇒ a=((2nπ±cos^(−1) ((√5)/5))/(ln 3)) x=((2nπ±cos^(−1) ((√5)/5))/(ln 3))−((ln 5)/(2ln 3))i; n∈Z

$$\mathrm{With}\:{x}={a}+{b}\mathrm{i};\:{a},\:{b}\:\in\mathbb{R}\:\mathrm{we}\:\mathrm{get}: \\ $$$$\mathrm{Real}\:\mathrm{part}: \\ $$$$\frac{\mathrm{1}}{\mathrm{9}^{{b}} }\left(\mathrm{2cos}^{\mathrm{2}} \:\left({a}\mathrm{ln}\:\mathrm{3}\right)\:−\mathrm{3}^{{b}} \mathrm{2cos}\:\left({a}\mathrm{ln}\:\mathrm{3}\right)\:+\mathrm{9}^{{b}} \mathrm{5}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{Imaginary}\:\mathrm{part}: \\ $$$$\frac{\mathrm{2}}{\mathrm{9}^{{b}} }\left(\mathrm{cos}\:\left({a}\mathrm{ln}\:\mathrm{3}\right)\:−\mathrm{3}^{{b}} \right)\mathrm{sin}\:\left({a}\mathrm{ln}\:\mathrm{3}\right)\:=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{9}^{{b}} }\neq\mathrm{0}\:\Rightarrow \\ $$$$\begin{cases}{\left(\mathrm{1}\right)\:\mathrm{cos}^{\mathrm{2}} \:\left({a}\mathrm{ln}\:\mathrm{3}\right)\:−\mathrm{3}^{{b}} \mathrm{cos}\:\left({a}\mathrm{ln}\:\mathrm{3}\right)\:+\frac{\mathrm{9}^{{b}} \mathrm{5}−\mathrm{1}}{\mathrm{2}}=\mathrm{0}}\\{\left(\mathrm{2}\right)\:\left(\mathrm{cos}\:\left({a}\mathrm{ln}\:\mathrm{3}\right)\:−\mathrm{3}^{{b}} \right)\mathrm{sin}\:\left({a}\mathrm{ln}\:\mathrm{3}\right)\:=\mathrm{0}}\end{cases} \\ $$$$\left(\mathrm{2}\right) \\ $$$$\mathrm{With}\:\mathrm{sin}\:\left({a}\mathrm{ln}\:\mathrm{3}\right)\:=\mathrm{0}\:\Rightarrow\:{b}\notin\mathbb{R} \\ $$$$\mathrm{cos}\:\left({a}\mathrm{ln}\:\mathrm{3}\right)\:=\mathrm{3}^{{b}} \:\Rightarrow \\ $$$$\left(\mathrm{1}\right)\:\mathrm{9}^{{b}} =\frac{\mathrm{1}}{\mathrm{5}}\:\Rightarrow\:{b}=−\frac{\mathrm{ln}\:\mathrm{5}}{\mathrm{2ln}\:\mathrm{3}} \\ $$$$\Rightarrow\:\mathrm{cos}\:\left({a}\mathrm{ln}\:\mathrm{3}\right)\:=\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\:\Rightarrow\:{a}=\frac{\mathrm{2}{n}\pi\pm\mathrm{cos}^{−\mathrm{1}} \:\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}}{\mathrm{ln}\:\mathrm{3}} \\ $$$$ \\ $$$${x}=\frac{\mathrm{2}{n}\pi\pm\mathrm{cos}^{−\mathrm{1}} \:\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}}{\mathrm{ln}\:\mathrm{3}}−\frac{\mathrm{ln}\:\mathrm{5}}{\mathrm{2ln}\:\mathrm{3}}\mathrm{i};\:{n}\in\mathbb{Z} \\ $$

Commented by Ghisom last updated on 20/Feb/24

thank you

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