17^x-1 =2e^2x+1
You asked:
Investigate the equation: \({17}^{x} - 1 = 2 \cdot {e}^{2} \cdot x + 1\).
MathBot Answer:
The 2 solutions to the equation are: \[\begin{aligned}x &= \frac{0.152126496633708365963753850572409800586897535395 - \frac{\log{\left(17 \right)}}{e^{2}}}{\log{\left(17 \right)}} \approx -0.081641305\\x &= \frac{3.19755622432931683950431224233020497311316547188 - \frac{\log{\left(17 \right)}}{e^{2}}}{\log{\left(17 \right)}} \approx 0.99326177\end{aligned}\]
\(e\) is Euler's number, a mathematical constant that is approximately \(2.71828\).