If x = log_{2a} a,y = log_{3a} 2a,and z = log | vedclass

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(A) Given $x = \log_{2a} a$,$y = \log_{3a} 2a$,and $z = \log_{4a} 3a$.Using the change of base formula $\log_b a = \frac{\ln a}{\ln b}$,we have:$x = \

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$1$

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(A) Given $x = \log_{2a} a$,$y = \log_{3a} 2a$,and $z = \log_{4a} 3a$.Using the change of base formula $\log_b a = \frac{\ln a}{\ln b}$,we have:$x = \frac{\ln a}{\ln 2a}$,$y = \frac{\ln 2a}{\ln 3a}$,$z = \frac{\ln 3a}{\ln 4a}$.Now,calculate $yz = \left(\frac{\ln 2a}{\ln 3a}\right) \left(\frac{\ln 3a}{\ln 4a}\right) = \frac{\ln 2a}{\ln 4a} = \log_{4a} 2a$.Next,calculate $xyz = \left(\frac{\ln a}{\ln 2a}\right) \left(\frac{\ln 2a}{\ln 3a}\right) \left(\frac{\ln 3a}{\ln 4a}\right) = \frac{\ln a}{\ln 4a} = \log_{4a} a$.Substitute these into the expression $yz(2-x) = 2yz - xyz$:$2(\log_{4a} 2a) - \log_{4a} a = \log_{4a} (2a)^2 - \log_{4a} a = \log_{4a} \left(\frac{4a^2}{a}\right) = \log_{4a} 4a = 1$.

If $x = \log_{2a} a$,$y = \log_{3a} 2a$,and $z = \log_{4a} 3a$,find the value of $yz(2-x)$.

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