ln(e^(25/c)-e^(5/c)+e^(-25/c)-e^(-5/c))
You asked:
Evaluate the expression: \(\ln\left( {e}^{\frac{25}{c}} - {e}^{\frac{5}{c}} + {e}^{\frac{-25}{c}} - {e}^{\frac{-5}{c}} \right)\)
MathBot Answer:
\[\ln\left( {e}^{\frac{25}{c}} - {e}^{\frac{5}{c}} + {e}^{\frac{-25}{c}} - {e}^{\frac{-5}{c}} \right) = \log{\left(e^{\frac{25}{c}} - e^{\frac{5}{c}} - e^{- \frac{5}{c}} + e^{- \frac{25}{c}} \right)}\]
\(e\) is Euler's number, a mathematical constant that is approximately \(2.71828\).