If x = log_{4/3}(1/2) and y = log_{1/2}(1/3), | vedclass

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(B) Given $x = \log_{4/3}(1/2)$. Since the base $4/3 > 1$ and the argument $1/2 < 1$,the value of $x$ is negative $(x < 0)$.Given $y = \log_{1/2}(1/3)

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$x < y$

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(B) Given $x = \log_{4/3}(1/2)$. Since the base $4/3 > 1$ and the argument $1/2 Given $y = \log_{1/2}(1/3)$. Using the property $\log_{a^n} b^m = \frac{m}{n} \log_a b$,we have $y = \log_{2^{-1}} (3^{-1}) = \frac{-1}{-1} \log_2 3 = \log_2 3$. Since the base $2 > 1$ and the argument $3 > 1$,the value of $y$ is positive $(y > 0)$.Comparing the two,since $x  0$,it follows that $x < y$.

If $x = \log_{4/3}(1/2)$ and $y = \log_{1/2}(1/3)$,then

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