ln(2)+0.5=t+e^-t
You asked:
Investigate the equation: \(\ln\left( 2 \right) + 0.5 = t + {e}^{-t}\).
MathBot Answer:
The 2 solutions to the equation are: \[\begin{aligned}t &= -1.2564312086261696769827376166092163269164168317 + \log{\left(2 \right)} \approx -0.56328403\\t &= -2.77555756156289135105907917022705078125 \cdot 10^{-17} + \log{\left(2 \right)} \approx 0.69314718\end{aligned}\]
\(e\) is Euler's number, a mathematical constant that is approximately \(2.71828\).