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

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(D) We simplify each term individually using the properties of logarithms: $\log_a b = 1/\log_b a$ and $a^{\log_a x} = x$.Term $1$: ${81^{(1/{\log_5}3

Vedclass · Accepted Answer

$890$

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(D) We simplify each term individually using the properties of logarithms: $\log_a b = 1/\log_b a$ and $a^{\log_a x} = x$.Term $1$: ${81^{(1/{\log_5}3)}} = {81^{\log_3 5}} = {(3^4)^{\log_3 5}} = {3^{4 \log_3 5}} = {3^{\log_3 5^4}} = {5^4} = 625$.Term $2$: ${27^{\log_9 36}} = {(3^3)^{\log_{3^2} 36}} = {3^{3 \cdot \frac{1}{2} \log_3 36}} = {3^{\frac{3}{2} \log_3 36}} = {3^{\log_3 36^{3/2}}} = {36^{3/2}} = {(6^2)^{3/2}} = 6^3 = 216$.Term $3$: ${3^{4/{\log_7}9}} = {3^{4 \log_9 7}} = {3^{4 \log_{3^2} 7}} = {3^{4 \cdot \frac{1}{2} \log_3 7}} = {3^{2 \log_3 7}} = {3^{\log_3 7^2}} = 7^2 = 49$.Summing the terms: $625 + 216 + 49 = 890$.

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

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