The value of {81^{(1/{log_5}3)}} + {27^{log_9 | vedclass

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(D) Given expression: ${81^{(1/{\log_5}3)}} + {27^{\log_9 36}} + {3^{4/{\log_7}9}}$Using the property $1/{\log_a b} = {\log_b a}$ and $n{\log_a b} = {

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$890$

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(D) Given expression: ${81^{(1/{\log_5}3)}} + {27^{\log_9 36}} + {3^{4/{\log_7}9}}$Using the property $1/{\log_a b} = {\log_b a}$ and $n{\log_a b} = {\log_a b^n}$:Term $1$: ${81^{\log_3 5}} = {(3^4)^{\log_3 5}} = {3^{4\log_3 5}} = {3^{\log_3 5^4}} = {5^4} = 625$Term $2$: ${27^{\log_9 36}} = {(3^3)^{\log_{3^2} 36}} = {3^{3 \cdot (1/2) \log_3 36}} = {3^{\log_3 36^{3/2}}} = {36^{3/2}} = {(6^2)^{3/2}} = {6^3} = 216$Term $3$: ${3^{4\log_9 7}} = {3^{4\log_{3^2} 7}} = {3^{4 \cdot (1/2) \log_3 7}} = {3^{2\log_3 7}} = {3^{\log_3 7^2}} = {7^2} = 49$Sum: $625 + 216 + 49 = 890$.

The value of ${81^{(1/{\log_5}3)}} + {27^{\log_9 36}} + {3^{4/{\log_7}9}}$ is equal to

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