7 log left( frac{16}{15} right) + 5 log left( | vedclass

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Question

(C) Let the expression be $E = 7 \log \left( \frac{16}{15} \right) + 5 \log \left( \frac{25}{24} \right) + 3 \log \left( \frac{81}{80} \right)$.Using

Vedclass · Accepted Answer

$\log 2$

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(C) Let the expression be $E = 7 \log \left( \frac{16}{15} ight) + 5 \log \left( \frac{25}{24} ight) + 3 \log \left( \frac{81}{80} ight)$.Using the property $n \log a = \log a^n$,we get:$E = \log \left( \left( \frac{16}{15} ight)^7 	imes \left( \frac{25}{24} ight)^5 	imes \left( \frac{81}{80} ight)^3 ight)$.Expressing the numbers in terms of prime factors:$16 = 2^4, 15 = 3 	imes 5, 25 = 5^2, 24 = 2^3 	imes 3, 81 = 3^4, 80 = 2^4 	imes 5$.Substituting these values:$E = \log \left( \frac{(2^4)^7}{(3 	imes 5)^7} 	imes \frac{(5^2)^5}{(2^3 	imes 3)^5} 	imes \frac{(3^4)^3}{(2^4 	imes 5)^3} ight)$.$E = \log \left( \frac{2^{28}}{3^7 	imes 5^7} 	imes \frac{5^{10}}{2^{15} 	imes 3^5} 	imes \frac{3^{12}}{2^{12} 	imes 5^3} ight)$.Grouping the powers of $2, 3,$ and $5$:$E = \log \left( \frac{2^{28}}{2^{15} 	imes 2^{12}} 	imes \frac{3^{12}}{3^7 	imes 3^5} 	imes \frac{5^{10}}{5^7 	imes 5^3} ight)$.$E = \log \left( \frac{2^{28}}{2^{27}} 	imes \frac{3^{12}}{3^{12}} 	imes \frac{5^{10}}{5^{10}} ight) = \log (2^1 	imes 1 	imes 1) = \log 2$.

$7 \log \left( \frac{16}{15} \right) + 5 \log \left( \frac{25}{24} \right) + 3 \log \left( \frac{81}{80} \right)$ is equal to

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