Question

Is there a solution to:
1/2 (e^x + e^-x) = 4

Answer 1

\[\Large \frac{1}{2}(e^x + e^{-x}) = 4\]
\[\Large e^x + e^{-x} = 4*2\]
\[\Large e^x + e^{-x} = 8\]
\[\Large e^x + \frac{1}{e^{x}} = 8\]
\[\Large e^{x}*(e^x + \frac{1}{e^{x}}) = 8e^{x}\]
\[\Large e^{x}*e^x + e^{x}\frac{1}{e^{x}} = 8e^{x}\]
\[\Large e^{2x} + 1 = 8e^{x}\]
\[\Large (e^{x})^2 + 1 = 8e^{x}\]
\[\Large z^2 + 1 = 8z\]
Solve the last equation (use the quadratic formula) then plug in z = e^x and solve for x

Answer 2

Awesome! Thank you!!

Answer 3

no problem