ln(0.01) = ln(1-e^(100t))
You asked:
Investigate the equation: \(\ln\left( 0.01 \right) = \ln\left( 1 - {e}^{100 t} \right)\).
MathBot Answer:
The real solution is: \[t = \frac{\log{\left(\frac{99}{100} \right)}}{100} \approx -0.00010050336\]
The complex solutions are: \[\left\{\frac{n i \pi}{50} + \frac{\log{\left(\frac{99}{100} \right)}}{100}\; \middle|\; n \in \mathbb{Z}\right\}\]
\(i\) is the imaginary unit, defined as \(i^2 = -1\).
\(e\) is Euler's number, a mathematical constant that is approximately \(2.71828\).