Question

If $a={\log }_{63}147$, $b={\log }_{189}1029$, then the value of $2\left(\frac{1-ab}{a-b}\right)$ is equal to

• A. $10$
• B. $63$
• C. $147$
• D. $60$

Answer 1

The correct option is A $10$

$a={\log }_{63}147(\because {\log }_{a}b=\frac{\log b}{\log a})$

$=\frac{\log 147}{\log 63}=\frac{\log (3\times 7^2)}{\log (3^2\times 7)}=\frac{\log 3+2\log 7}{2\log 3+\log 7}[\because \log (ab)=\log a+\log b]$

$b={\log }_{189}1029=\frac{\log 1029}{\log 189}=\frac{\log \left({3\times 7}^3\right)}{\log (3^3\times 7)}=\frac{\log 3+3\log 7}{3\log 3+\log 7}$

$\therefore 2\left(\frac{1-ab}{a-b}\right)=2\left(\frac{1-\frac{\log 3+2\log 7}{2\log 3+\log 7}\times \frac{\log 3+3\log 7}{3\log 3+\log 7}}{\frac{\log 3+2\log 7}{2\log 3+\log 7}-\frac{\log 3+3\log 7}{3\log 3+\log 7}}\right)$

$=2\left(\frac{(2\log 3+\log 7)(3\log 3+\log 7)-(\log 3+2\log 7)(\log 3+3\log 7)}{(\log 3+2\log 7)(3\log 3+\log 7)-(\log 3+3\log 7)(2\log 3+\log 7)}\right)$

$=2\left(\frac{6{\left(\log 3\right)}^2+5\log 3\log 7+{\left(\log 7\right)}^2-{\left(\log 3\right)}^2-5\log 3\log 7-6{\left(\log 7\right)}^2}{3{\left(\log 3\right)}^2+7\log 3\log 7+2{\left(\log 7\right)}^2-2{\left(\log 3\right)}^2-7\log 3\log 7-3{\left(\log 7\right)}^2}\right)$

$=2\left(\frac{5{\left(\log 3\right)}^2-5{\left(\log 7\right)}^2}{{\left(\log 3\right)}^2-{\left(\log 7\right)}^2}\right)=2\left(5\right)=10$

Hence, option A.