frac{log 49 sqrt{7} + log 25 sqrt{5} - log 4 | vedclass

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(C) The expression is $\frac{\log(49 \cdot 7^{1/2}) + \log(25 \cdot 5^{1/2}) - \log(4 \cdot 2^{1/2})}{\log 17.5}$.Since $49 = 7^2$,$25 = 5^2$,and $4 =

Vedclass · Accepted Answer

$\frac{5}{2}$

Vedclass · Answer

(C) The expression is $\frac{\log(49 \cdot 7^{1/2}) + \log(25 \cdot 5^{1/2}) - \log(4 \cdot 2^{1/2})}{\log 17.5}$.Since $49 = 7^2$,$25 = 5^2$,and $4 = 2^2$,we have:$\log(7^2 \cdot 7^{1/2}) = \log(7^{5/2}) = \frac{5}{2} \log 7$.$\log(5^2 \cdot 5^{1/2}) = \log(5^{5/2}) = \frac{5}{2} \log 5$.$\log(2^2 \cdot 2^{1/2}) = \log(2^{5/2}) = \frac{5}{2} \log 2$.Substituting these into the numerator:$\frac{5}{2} \log 7 + \frac{5}{2} \log 5 - \frac{5}{2} \log 2 = \frac{5}{2} (\log 7 + \log 5 - \log 2)$.Using log properties $\log a + \log b = \log(ab)$ and $\log a - \log b = \log(a/b)$:$\log 7 + \log 5 - \log 2 = \log(\frac{7 \cdot 5}{2}) = \log(\frac{35}{2}) = \log 17.5$.Thus,the expression becomes:$\frac{\frac{5}{2} \log 17.5}{\log 17.5} = \frac{5}{2}$.

$\frac{\log 49 \sqrt{7} + \log 25 \sqrt{5} - \log 4 \sqrt{2}}{\log 17.5} = $

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