7 log frac{16}{15} + 5 log frac{25}{24} + 3 l | vedclass

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(A) We use the property $n \log a = \log a^n$ and $\log(a/b) = \log a - \log b$.First,express the numbers in prime factors:$7 \log \left(\frac{2^4}{3

Vedclass · Accepted Answer

$\log 2$

Vedclass · Answer

(A) We use the property $n \log a = \log a^n$ and $\log(a/b) = \log a - \log b$.First,express the numbers in prime factors:$7 \log \left(\frac{2^4}{3 	imes 5}ight) + 5 \log \left(\frac{5^2}{2^3 	imes 3}ight) + 3 \log \left(\frac{3^4}{2^4 	imes 5}ight)$$= 7(4 \log 2 - \log 3 - \log 5) + 5(2 \log 5 - 3 \log 2 - \log 3) + 3(4 \log 3 - 4 \log 2 - \log 5)$$= (28 \log 2 - 7 \log 3 - 7 \log 5) + (10 \log 5 - 15 \log 2 - 5 \log 3) + (12 \log 3 - 12 \log 2 - 3 \log 5)$Now,group the terms by $\log 2, \log 3,$ and $\log 5$:$\log 2: 28 - 15 - 12 = 1$$\log 3: -7 - 5 + 12 = 0$$\log 5: -7 + 10 - 3 = 0$Result $= 1 \log 2 + 0 \log 3 + 0 \log 5 = \log 2$.

$7 \log \frac{16}{15} + 5 \log \frac{25}{24} + 3 \log \frac{81}{80} =$

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