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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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,$E = \frac{3 \log_{10} 70}{\log_{10} 70} = 3$.

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

Similar Questions

If $x = \log_{4/3}(1/2)$ and $y = \log_{1/2}(1/3)$,then

If $3^{x-2}=5$ and $\log_{10} 2=0.30103, \log_{10} 3=0.4771$,then $x=$

If $\log _{2} x+\log _{4} x+\log _{16} x=21 / 4,$ then $x=$

If $\frac{\log x}{2} = \frac{\log y}{3} = \frac{\log z}{5}$,then $yz$ in terms of $x$ is

$\log _{5}\left(1+\frac{1}{5}\right)+\log _{5}\left(1+\frac{1}{6}\right)+\log _{5}\left(1+\frac{1}{7}\right)+\cdots+\log _{5} \left(1+\frac{1}{624}\right)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module