Question

If ${\log }_{10}15=a$, ${\log }_{20}50=b$, then the value of $\frac{5-b}{2ab+2a-4b+2}$ is equal to

• A. ${\log }_{9}40$
• B. ${\log }_{9}35$
• C. ${\log }_{9}9$
• D. ${\log }_{40}40$

Answer 1

The correct option is A ${\log }_{9}40$

Given ${\log }_{10}15=a$ and ${\log }_{20}50=b$ ...(i)
$\frac{5-b}{2ab+2a-4b+2}$ = $\frac{5-{\log }_{20}50}{(2{\log }_{10}15\times {\log }_{20}50)+2{\log }_{10}15-4{\log }_{20}50+2}$ (using (1))
$=\frac{5-\frac{{\log }_{e}50}{{\log }_{e}20}}{2\left(\frac{{\log }_{e}15}{{\log }_{e}10}\right)\left(\frac{{\log }_{e}50}{{\log }_{e}20}\right)+2\left(\frac{{\log }_{e}15}{{\log }_{e}10}\right)-4\left(\frac{{\log }_{e}50}{{\log }_{e}20}\right)+2}$ ($\because {\log }_{a}b=\frac{{\log }_{e}b}{{\log }_{e}a}$)
$=\frac{\frac{5{\log }_{e}20-{\log }_{e}50}{{\log }_{e}20}}{\frac{2{\log }_{e}15{\log }_{e}50+2{\log }_{e}15{\log }_{e}20-4{\log }_{e}10{\log }_{e}50+2{\log }_{e}10{\log }_{e}20}{{\log }_{e}10\times {\log }_{e}20}}$
$=\frac{{\log }_{e}10\left[{\log }_e\right(20{)}^5-{\log }_{e}50]}{2{\log }_{e}15({\log }_{e}50+{\log }_{e}20)+2{\log }_{e}10({\log }_{e}20-2{\log }_{e}50)}$ ($\because \log x^m=m\log x$)
$=\frac{{\log }_{e}10\left[{\log }_e\right(20{)}^5-{\log }_{e}50]}{{\log }_e\left(15{)}^2\right({\log }_{e}1000)+2{\log }_{e}10[{\log }_{e}20-{\log }_e\left(50{)}^2\right]}$ ($\because \log \left(mn\right)=\log m+\log n$)
$=\frac{{\log }_{e}10\left[{\log }_e\frac{(20{)}^5}{50}\right]}{{\log }_{e}10[3{\log }_{e}225+2{\log }_e\frac{20}{(50{)}^2}]}$ ($\because \log \left(\frac{m}{n}\right)=\log m-\log n$)
$=\frac{{\log }_e\frac{(20{)}^5}{50}}{{\log }_e(225{)}^3+{\log }_e{\left(\frac{20}{(50{)}^2}\right)}^2}$
$=\frac{{\log }_e\frac{(20{)}^5}{50}}{{\log }_e\left[\right(225{)}^3\frac{(20{)}^2}{(50{)}^4}]}$ ($\because \log \left(mn\right)=\log m+\log n$)
$=\frac{{\log }_e(40{)}^3}{{\log }_e(9{)}^3}$
$=\frac{3{\log }_{e}40}{3{\log }_{e}9}$ ($\because {\log }_{a}b=\frac{{\log }_{e}b}{{\log }_{e}a}$)
$=\frac{{\log }_{e}40}{{\log }_{e}9}={\log }_{9}40$