Question 14 Marks
Find the value of $\frac{3+\log _{10} 343}{2+\frac{1}{2} \log _{10}\left(\frac{49}{4}\right)+\frac{1}{2} \log _{10}\left(\frac{1}{25}\right)}$
Answer
View full question & answer→$ \frac{3+\log _{10} 343}{2+\frac{1}{2} \log _{10}\left(\frac{49}{4}\right)+\frac{1}{2} \log _{10}\left(\frac{1}{25}\right)}$
$=\frac{3+\log _{10} 7^3}{2+\log _{10}\left(\frac{49}{4}\right)^{\frac{1}{2}}+\log _{10}\left(\frac{1}{25}\right)^{\frac{1}{2}}}$
$=\frac{3+3 \cdot \log _{10} 7}{2+\log _{10} \frac{7}{2}+\log _{10} \frac{1}{5}}$
$=\frac{3\left(1+\log _{10} 7\right)}{2+\log _{10}\left(\frac{7}{2} \times \frac{1}{5}\right)}$
$=\frac{3\left(1+\log _{10} 7\right)}{2+\log _{10}\left(\frac{7}{10}\right)}$
$=\frac{3\left(1+\log _{10} 7\right)}{2+\log _{10} 7-\log _{10} 10}$
$=\frac{3\left(1+\log _{10} 7\right)}{2+\log _{10} 7-1}$
$\ldots\left[\because \log _a a=1\right]$
$=\frac{3\left(1+\log _{10} 7\right)}{1+\log _{10} 7}=3$
$=\frac{3+\log _{10} 7^3}{2+\log _{10}\left(\frac{49}{4}\right)^{\frac{1}{2}}+\log _{10}\left(\frac{1}{25}\right)^{\frac{1}{2}}}$
$=\frac{3+3 \cdot \log _{10} 7}{2+\log _{10} \frac{7}{2}+\log _{10} \frac{1}{5}}$
$=\frac{3\left(1+\log _{10} 7\right)}{2+\log _{10}\left(\frac{7}{2} \times \frac{1}{5}\right)}$
$=\frac{3\left(1+\log _{10} 7\right)}{2+\log _{10}\left(\frac{7}{10}\right)}$
$=\frac{3\left(1+\log _{10} 7\right)}{2+\log _{10} 7-\log _{10} 10}$
$=\frac{3\left(1+\log _{10} 7\right)}{2+\log _{10} 7-1}$
$\ldots\left[\because \log _a a=1\right]$
$=\frac{3\left(1+\log _{10} 7\right)}{1+\log _{10} 7}=3$