ln(2)+0.5=t+e^-t

asked by guest
on Nov 10, 2024 at 2:08 pm



You asked:

Investigate the equation: ln⁡(2)+0.5=t+e−t\ln\left( 2 \right) + 0.5 = t + {e}^{-t}.

MathBot Answer:

The 2 solutions to the equation are: t=−1.2564312086261696769827376166092163269164168317+log⁡(2)≈−0.56328403t=−2.77555756156289135105907917022705078125⋅10−17+log⁡(2)≈0.69314718\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}

ee is Euler's number, a mathematical constant that is approximately 2.718282.71828.

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