(ln(e^y))^2=

asked by guest
on Jan 22, 2025 at 10:21 am



You asked:

Evaluate the expression: \({\ln\left( {e}^{y} \right)}^{2}\)

MathBot Answer:

\[{\ln\left( {e}^{y} \right)}^{2} = \log{\left(e^{y} \right)}^{2}\]


\(e\) is Euler's number, a mathematical constant that is approximately \(2.71828\).