解決operatornamede^J=sqrt[3]{10^29/10^5} |Microsoft數學求解器

$$\operatorname { de } ^ { J } = \sqrt[ 3 ] { \frac { 10 ^ { 29 } } { 10 ^ { 5 } } }$$

解 J (復數求解)

$$J=\frac{2\pi n_{1}i}{\ln(d)+1}+\frac{8\ln(10)}{\ln(d)+1}$$

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

$$d\neq \frac{1}{e}\text{ and }d\neq 0$$

解 d (復數求解)

$$d=e^{-\frac{2\pi n_{1}iRe(J)}{\left(Re(J)\right)^{2}+\left(Im(J)\right)^{2}}+\frac{-\left(Re(J)\right)^{2}-\left(Im(J)\right)^{2}}{\left(Re(J)\right)^{2}+\left(Im(J)\right)^{2}}-\frac{2\pi n_{1}Im(J)}{\left(Re(J)\right)^{2}+\left(Im(J)\right)^{2}}}\times 100000000^{\frac{Re(J)}{\left(Re(J)\right)^{2}+\left(Im(J)\right)^{2}}}\times 10^{-\frac{8iIm(J)}{\left(Re(J)\right)^{2}+\left(Im(J)\right)^{2}}}$$

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

$$\left(Re(J)\right)^{2}+\left(Im(J)\right)^{2}\neq 0$$

解 J

$$J=\frac{8\ln(10)}{\ln(d)+1}$$

$$d\neq \frac{1}{e}\text{ and }d>0$$

解 d

$$\left\{\begin{matrix}d=-\frac{100000000^{\frac{1}{J}}}{e}\text{, }&J\neq 0\text{ and }Numerator(J)\text{bmod}2=0\text{ and }Denominator(J)\text{bmod}2=1\\d=\frac{100000000^{\frac{1}{J}}}{e}\text{, }&J\neq 0\end{matrix}\right.$$