frac{3+log _{10} 343}{2+frac{1}{2} log _{10} | vedclass

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Question

(A) Given expression: $E = \frac{3 + \log_{10} 343}{2 + \frac{1}{2} \log_{10} \left(\frac{49}{4}\right) + \frac{1}{3} \log_{10} \left(\frac{1}{125}\ri

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) Given expression: $E = \frac{3 + \log_{10} 343}{2 + \frac{1}{2} \log_{10} \left(\frac{49}{4}ight) + \frac{1}{3} \log_{10} \left(\frac{1}{125}ight)}$Numerator: $3 + \log_{10} 7^3 = 3 + 3 \log_{10} 7 = 3(1 + \log_{10} 7) = 3(\log_{10} 10 + \log_{10} 7) = 3 \log_{10} 70$.Denominator: $2 + \log_{10} \left(\frac{49}{4}ight)^{1/2} + \log_{10} \left(\frac{1}{125}ight)^{1/3} = 2 + \log_{10} \left(\frac{7}{2}ight) + \log_{10} \left(\frac{1}{5}ight)$.Using $\log a + \log b = \log(ab)$,the denominator becomes: $2 + \log_{10} \left(\frac{7}{2} 	imes \frac{1}{5}ight) = 2 + \log_{10} \left(\frac{7}{10}ight)$.Since $2 = \log_{10} 100$,the denominator is: $\log_{10} 100 + \log_{10} 0.7 = \log_{10} (100 	imes 0.7) = \log_{10} 70$.Thus,$E = \frac{3 \log_{10} 70}{\log_{10} 70} = 3$.

$\frac{3+\log _{10} 343}{2+\frac{1}{2} \log _{10} \left(\frac{49}{4}\right)+\frac{1}{3} \log _{10} \left(\frac{1}{125}\right)}=$

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