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(B) The given equation is ${x^2} - 3kx + 2{e^{2\log k}} - 1 = 0$.Using the property of logarithms,${e^{2\log k}} = {e^{\log {k^2}}} = {k^2}$.Thus,the

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$k = 2$

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(B) The given equation is ${x^2} - 3kx + 2{e^{2\log k}} - 1 = 0$.Using the property of logarithms,${e^{2\log k}} = {e^{\log {k^2}}} = {k^2}$.Thus,the equation becomes ${x^2} - 3kx + 2{k^2} - 1 = 0$.Since the product of the roots is $7$,we have $2{k^2} - 1 = 7$,which implies $2{k^2} = 8$,so ${k^2} = 4$,giving $k = \pm 2$.Since the term $\log k$ is present in the original equation,$k$ must be positive,so $k = 2$.For the roots to be real,the discriminant $D = b^2 - 4ac$ must be $\ge 0$.Here,$D = (-3k)^2 - 4(1)(2{k^2} - 1) = 9{k^2} - 8{k^2} + 4 = {k^2} + 4$.Since ${k^2} + 4$ is always greater than $0$ for any real $k$,the roots are real for $k = 2$.

If the product of the roots of the equation ${x^2} - 3kx + 2{e^{2\log k}} - 1 = 0$ is $7$,then for what value of $k$ will its roots be real?

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