15*10^(-9)=5*10^(-19)(e^((x)/(0.052))-1)+10*10^(-19)(e^((x)/(0.052))-1)
You asked:
Investigate the equation: \(15 \cdot {10}^{-9} = 5 \cdot {10}^{-19} \cdot \left({e}^{\frac{x}{0.052}} - 1\right) + 10 \cdot {10}^{-19} \cdot \left({e}^{\frac{x}{0.052}} - 1\right)\).
MathBot Answer:
The real solution is: \[x = \frac{13 \log{\left(10000000001 \right)}}{250} \approx 1.1973442\]
The complex solutions are: \[\left\{\frac{13 n i \pi}{125} + \frac{13 \log{\left(10000000001 \right)}}{250}\; \middle|\; n \in \mathbb{Z}\right\}\]
\(i\) is the imaginary unit, defined as \(i^2 = -1\).
\(e\) is Euler's number, a mathematical constant that is approximately \(2.71828\).