13000=5000\left(1+\frac{x}{1}\right)^{12}

asked by guest
on Nov 28, 2024 at 6:54 pm



You asked:

Solve the equation \(13000 = 5000 {\left( 1 + \frac{x}{1} \right)}^{12}\) for the variable \(x\).

MathBot Answer:

The solutions are given by the formula: \[x = \left(-1 + \frac{5^{\frac{11}{12}} \cdot \sqrt[12]{13} \cos{\left(\frac{\pi k}{6} \right)}}{5}\right) + \frac{5^{\frac{11}{12}} \cdot \sqrt[12]{13} \sin{\left(\frac{\pi k}{6} \right)}}{5} i\] where \(k\) is an integer in \(0 \le k < 12\).

There are \(2\) real solutions.

There are \(10\) solutions with nonzero imaginary part.


Real solutions: \[\begin{aligned}x &= -1 + \frac{5^{\frac{11}{12}} \cdot \sqrt[12]{13}}{5} \\&\approx 0.082881944\\x &= -1 - \frac{5^{\frac{11}{12}} \cdot \sqrt[12]{13}}{5} \\&\approx -2.0828819\end{aligned}\]


Solutions with nonzero imaginary part (\(8\) of \(10\) displayed): \[\begin{aligned}x &= -1 + \frac{\sqrt[12]{13} \sqrt{3} \cdot 5^{\frac{11}{12}}}{10} + \frac{\sqrt[12]{13} \cdot 5^{\frac{11}{12}} i}{10} \\&\approx -0.062196727 + 0.54144097 i\\x &= -1 + \frac{\sqrt[12]{13} \cdot 5^{\frac{11}{12}}}{10} + \frac{\sqrt[12]{13} \sqrt{3} \cdot 5^{\frac{11}{12}} i}{10} \\&\approx -0.45855903 + 0.93780327 i\\x &= -1 + \frac{\sqrt[12]{13} \cdot 5^{\frac{11}{12}} i}{5} \\&= -1 + 1.0828819 i\\x &= -1 - \frac{\sqrt[12]{13} \cdot 5^{\frac{11}{12}}}{10} + \frac{\sqrt[12]{13} \sqrt{3} \cdot 5^{\frac{11}{12}} i}{10} \\&\approx -1.541441 + 0.93780327 i\\x &= -1 - \frac{\sqrt[12]{13} \sqrt{3} \cdot 5^{\frac{11}{12}}}{10} + \frac{\sqrt[12]{13} \cdot 5^{\frac{11}{12}} i}{10} \\&\approx -1.9378033 + 0.54144097 i\\x &= -1 - \frac{\sqrt[12]{13} \sqrt{3} \cdot 5^{\frac{11}{12}}}{10} - \frac{\sqrt[12]{13} \cdot 5^{\frac{11}{12}} i}{10} \\&\approx -1.9378033 -0.54144097 i\\x &= -1 - \frac{\sqrt[12]{13} \cdot 5^{\frac{11}{12}}}{10} - \frac{\sqrt[12]{13} \sqrt{3} \cdot 5^{\frac{11}{12}} i}{10} \\&\approx -1.541441 -0.93780327 i\\x &= -1 - \frac{\sqrt[12]{13} \cdot 5^{\frac{11}{12}} i}{5} \\&= -1 -1.0828819 i\end{aligned}\]


\(i\) is the imaginary unit, defined as \(i^2 = -1\).