If the product of the roots of the equation $\displaystyle x^{2}-5x+4^{log_{2}\lambda }=0$ is $8$, then $\lambda $ is

By Vieta's formulas, the product of the roots is the constant term, so we set $4^{\log_{2}\lambda} = 8$. Applying the base change formula to the bases, we rewrite 4 as $2^2$ and 8 as $2^3$, giving $2^{2\log_{2}\lambda} = 2^3$. By the properties of logarithms, this simplifies to $\lambda^2 = 8$. Solving for the positive real value of $\lambda$, we find $\lambda = 2\sqrt{2}$.

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