समाधान 2^n+3+ilog_{2}x-8=(1-log_{2}y)i

$$2 ^ { n + 3 } + i \log _ { 2 } x - 8 = ( 1 - \log _ { 2 } y ) i$$

n को लागि हल गर्नुहोस्

$$n=\frac{\ln(\frac{-i\ln(x)-i\ln(y)+\ln(2)\left(8+i\right)}{\ln(2)})-3\ln(2)}{\ln(2)}+\frac{2\pi n_{1}i}{\ln(2)}$$

$$n_{1}\in \mathrm{Z}$$

$$\left(Im(\ln(y))+Im(\ln(\frac{2^{1-8i}}{y}))+8\ln(2)\neq 0\text{ and }x\neq 0\text{ and }y\neq 0\text{ and }Im(\ln(y))+Im(\ln(\frac{2^{1-8i}}{y}))\neq -8\ln(2)\right)\text{ or }\left(x\neq \frac{2^{1-8i}}{y}\text{ and }x\neq 0\text{ and }y\neq 0\right)$$

x को लागि हल गर्नुहोस्

$$x=\frac{2^{8i\times 2^{n}+\left(1-8i\right)}}{y}$$

$$Im(\ln(\frac{2^{8i\times 2^{n}+\left(1-8i\right)}}{y}))-8\ln(2)Re(2^{n})+Im(\ln(y))+8\ln(2)=0\text{ and }y\neq 0$$