17^(x-1) =2e^(2x+1)
You asked:
Investigate the equation: \({17}^{x - 1} = 2 \cdot {e}^{2 x + 1}\).
MathBot Answer:
The solution is: \[x = \frac{1 + \log{\left(2 \right)} + \log{\left(17 \right)}}{-2 + \log{\left(17 \right)}} \approx 5.4324148\]
\(e\) is Euler's number, a mathematical constant that is approximately \(2.71828\).