Plug x=(1/4)ln(k) into ...?
g(x) = e^(2x) + k*e^(-2x)
The answer is 2*sqrt(k) but I don’t see how?
回答 (4)
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.
收錄日期: 2021-04-18 18:29:05
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