Question and Answers Forum

All Questions      Topic List

Logarithms Questions

Previous in All Question      Next in All Question      

Previous in Logarithms      Next in Logarithms      

Question Number 91608 by mhmd last updated on 01/May/20

Commented by Tony Lin last updated on 01/May/20

$$let{log}_{2}x=t$$$$\frac{1}{4}{t}^2-t+4=0$$$$t^2-4t+16=0$$$$t=\frac{4\pm 48i}{2}=2\pm 24i$$$${log}_{2}x=2\pm 24i$$$$x=2^{2\pm 24i}=4\times 2^{\pm 24i}$$$$=4\times e^{\pm 24ln2i}$$$$=4\times [cos\left(24ln2\right)\pm isin\left(24ln2\right)]$$$$\approx 4\times (-0.6\pm 0.8i)$$$$=-2.4\pm 3.2i$$

Commented by abdomathmax last updated on 02/May/20

$$\left(e\right)\to {\left(\frac{ln\left(x\right)}{2ln2}\right)}^2-\frac{ln\left(x\right)}{ln\left(2\right)}+4=0\Rightarrow$$$$\frac{{ln}^2\left(x\right)}{4{ln}^2\left(2\right)}-\frac{lnx}{ln\left(2\right)}+4=0\Rightarrow$$$$4{ln}^2\left(2\right)\{\frac{{ln}^2\left(x\right)}{4{ln}^{2)}2}-\frac{lnx}{ln2}+4\}=0\Rightarrow$$$${ln}^2\left(x\right)-4ln\left(2\right)ln\left(x\right)+16{ln}^2\left(2\right)=0$$$$letln\left(x\right)=t\Rightarrow t^2-4ln\left(2\right)t+16{ln}^2\left(2\right)=0$$$${\mathrm{\Delta }}^{{}^{\prime }}=4{ln}^{2}2-16{ln}^{2}2=-12{ln}^2\left(2\right)={\left(2i\sqrt{3}ln\left(2\right)\right)}^2$$$$t_1=2ln\left(2\right)+2i\sqrt{3}ln\left(2\right)=2ln\left(2\right)(1+i\sqrt{3})$$$$=4ln\left(2\right)e^{\frac{i\pi }{3}}and{t}_2=4ln\left(2\right)e^{-\frac{i\pi }{3}}$$$$x_1=e^{t_1}=e^{2ln\left(2\right)}{e}^{i\left(2\sqrt{3}ln2\right)}$$$$=4(cos\left(2\sqrt{3}ln\left(2\right)\right)+isin\left(2\sqrt{3}ln\left(2\right)\right)$$$$x_2=4\{cos\left(2\sqrt{3}ln2\right)-isin\left(2\sqrt{3}ln\left(2\right)\right\}$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com