If log _{8} p=2.5 and log _{2} q=5,then p in | vedclass

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(A) Given $\log _{8} p = 2.5 = \frac{5}{2}$.Using the definition of logarithm,$p = 8^{5/2}$.Since $8 = 2^3$,we have $p = (2^3)^{5/2} = 2^{15/2}$.Given

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$q \sqrt{q}$

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(A) Given $\log _{8} p = 2.5 = \frac{5}{2}$.Using the definition of logarithm,$p = 8^{5/2}$.Since $8 = 2^3$,we have $p = (2^3)^{5/2} = 2^{15/2}$.Given $\log _{2} q = 5$,we have $q = 2^5$.We can express $p$ in terms of $q$ by substituting $2^5 = q$ into the expression for $p$:$p = 2^{15/2} = (2^5)^{3/2} = q^{3/2}$.Since $q^{3/2} = q^1 \cdot q^{1/2} = q\sqrt{q}$,the correct option is $A$.

If $\log _{8} p=2.5$ and $\log _{2} q=5$,then $p$ in terms of $q$ is

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