Plug x=(1/4)ln(k) into ...?

The answer is as follows:

term at a time
e^(2x) = e^(2(1/4)ln(k) ) = e^((1/2)ln(k) ) = e^(ln(√k) ) = √k

k•e^(-2(1/4)ln(k) ) = k•e^((–1/2)ln(k) )  = k•e^(ln(1/√k) ) = k/√k
   = (k/√k) (√k/√k) = √k

add the two to get 2√k

we have 1/4ln(k)=ln(k^(1/4)) and e^ln(x)=x so the answer is 2*sqrt(k)

g(x) = e^(2x) + k*e^(-2x)
g(1/4)ln(k) = e^(2x) + k/e^(2x) =>
(e^(4x) + k ) /e^(2x) =>
(e^(4(1/4)ln(k))) + k ) / e^(2(1/4)ln(k))
(e^(ln(k) + k) / e^(1/2ln(k)) =>

Remember 'e^ln = 1'
Hence
(k + k) / e^(ln(k^(1/2))
2k / k^(1/2) =>
2k^(1/2) ; 2sqrt(k) As required.