1,000(1+x)^{12}=4,300

asked by guest
on Jan 27, 2025 at 9:00 am



You asked:

Solve the equation \(1,000 {\left( 1 + x \right)}^{12} = 4,300\) for the variable \(x\).

MathBot Answer:

The solutions are given by the formula: \[x = \left(-1 + \frac{10^{\frac{11}{12}} \cdot \sqrt[12]{43} \cos{\left(\frac{\pi k}{6} \right)}}{10}\right) + \frac{10^{\frac{11}{12}} \cdot \sqrt[12]{43} \sin{\left(\frac{\pi k}{6} \right)}}{10} i\] 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: \[\begin{aligned}x &= -1 + \frac{10^{\frac{11}{12}} \cdot \sqrt[12]{43}}{10} \\&\approx 0.12924724\\x &= -1 - \frac{10^{\frac{11}{12}} \cdot \sqrt[12]{43}}{10} \\&\approx -2.1292472\end{aligned}\]


Solutions with nonzero imaginary part (\(8\) of \(10\) displayed): \[\begin{aligned}x &= -1 + \frac{10^{\frac{11}{12}} \sqrt{3} \cdot \sqrt[12]{43}}{20} + \frac{10^{\frac{11}{12}} \cdot \sqrt[12]{43} i}{20} \\&\approx -0.022043203 + 0.56462362 i\\x &= -1 + \frac{10^{\frac{11}{12}} \cdot \sqrt[12]{43}}{20} + \frac{10^{\frac{11}{12}} \sqrt{3} \cdot \sqrt[12]{43} i}{20} \\&\approx -0.43537638 + 0.9779568 i\\x &= -1 + \frac{10^{\frac{11}{12}} \cdot \sqrt[12]{43} i}{10} \\&= -1 + 1.1292472 i\\x &= -1 - \frac{10^{\frac{11}{12}} \cdot \sqrt[12]{43}}{20} + \frac{10^{\frac{11}{12}} \sqrt{3} \cdot \sqrt[12]{43} i}{20} \\&\approx -1.5646236 + 0.9779568 i\\x &= -1 - \frac{10^{\frac{11}{12}} \sqrt{3} \cdot \sqrt[12]{43}}{20} + \frac{10^{\frac{11}{12}} \cdot \sqrt[12]{43} i}{20} \\&\approx -1.9779568 + 0.56462362 i\\x &= -1 - \frac{10^{\frac{11}{12}} \sqrt{3} \cdot \sqrt[12]{43}}{20} - \frac{10^{\frac{11}{12}} \cdot \sqrt[12]{43} i}{20} \\&\approx -1.9779568 -0.56462362 i\\x &= -1 - \frac{10^{\frac{11}{12}} \cdot \sqrt[12]{43}}{20} - \frac{10^{\frac{11}{12}} \sqrt{3} \cdot \sqrt[12]{43} i}{20} \\&\approx -1.5646236 -0.9779568 i\\x &= -1 - \frac{10^{\frac{11}{12}} \cdot \sqrt[12]{43} i}{10} \\&= -1 -1.1292472 i\end{aligned}\]


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