25=(1+r/100)^9

asked by guest
on Jan 01, 2025 at 1:57 am



You asked:

Solve the equation \(25 = {\left( 1 + \frac{r}{100} \right)}^{9}\) for the variable \(r\).

MathBot Answer:

The solutions are given by the formula: \[r = \left(-100 + 100 \cdot 5^{\frac{2}{9}} \cos{\left(\frac{2 \pi k}{9} \right)}\right) + 100 \cdot 5^{\frac{2}{9}} \sin{\left(\frac{2 \pi k}{9} \right)} i\] where \(k\) is an integer in \(0 \le k < 9\).

There is \(1\) real solution.

There are \(8\) solutions with nonzero imaginary part.


Real solutions: \[\begin{aligned}r &= -100 + 100 \cdot 5^{\frac{2}{9}} \\&\approx 42.996915\end{aligned}\]


Solutions with nonzero imaginary part: \[\begin{aligned}r &= -100 + 100 \cdot 5^{\frac{2}{9}} \cos{\left(\frac{2 \pi}{9} \right)} + 100 \cdot 5^{\frac{2}{9}} i \sin{\left(\frac{2 \pi}{9} \right)} \\&\approx 9.541992 + 91.916645 i\\r &= -100 + 100 \cdot 5^{\frac{2}{9}} \cos{\left(\frac{4 \pi}{9} \right)} + 100 \cdot 5^{\frac{2}{9}} i \sin{\left(\frac{4 \pi}{9} \right)} \\&\approx -75.168846 + 140.82447 i\\r &= -100 - 50 \cdot 5^{\frac{2}{9}} + 50 \sqrt{3} \cdot 5^{\frac{2}{9}} i \\&\approx -171.49846 + 123.83896 i\\r &= - 100 \cdot 5^{\frac{2}{9}} \cos{\left(\frac{\pi}{9} \right)} - 100 + 100 \cdot 5^{\frac{2}{9}} i \sin{\left(\frac{\pi}{9} \right)} \\&\approx -234.37315 + 48.907825 i\\r &= - 100 \cdot 5^{\frac{2}{9}} \cos{\left(\frac{\pi}{9} \right)} - 100 - 100 \cdot 5^{\frac{2}{9}} i \sin{\left(\frac{\pi}{9} \right)} \\&\approx -234.37315 -48.907825 i\\r &= -100 - 50 \cdot 5^{\frac{2}{9}} - 50 \sqrt{3} \cdot 5^{\frac{2}{9}} i \\&\approx -171.49846 -123.83896 i\\r &= -100 + 100 \cdot 5^{\frac{2}{9}} \cos{\left(\frac{4 \pi}{9} \right)} - 100 \cdot 5^{\frac{2}{9}} i \sin{\left(\frac{4 \pi}{9} \right)} \\&\approx -75.168846 -140.82447 i\\r &= -100 + 100 \cdot 5^{\frac{2}{9}} \cos{\left(\frac{2 \pi}{9} \right)} - 100 \cdot 5^{\frac{2}{9}} i \sin{\left(\frac{2 \pi}{9} \right)} \\&\approx 9.541992 -91.916645 i\end{aligned}\]


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