Question

If ${\log }_{245}175=a,{\log }_{1715}875=b$ then find the value of $\frac{1-ab}{a-b}$.

Answer 1

Answer: 5

As, $175=5^2\times 7;245=5\times 7^2;875=5^3\times 7;1715=5\times 7^3$
Now, ${\log }_{\left(5\times 7^2\right)}\left(5^2\times 7\right)=a\Rightarrow \frac{2+{\log }_{5}7}{1+2{\log }_{5}7}=a$ ...(1)
Also, ${\log }_{\left(5\times 7^3\right)}\left(5^3\times 7\right)=a\Rightarrow \frac{3+{\log }_{5}7}{1+3{\log }_{5}7}=b$ ....(2)
$\therefore$ from(1), we get
$2+{\log }_{5}7=a+2a\cdot {\log }_{5}7\Rightarrow (2-a)=(2a-1){\log }_{5}7\Rightarrow {\log }_{5}7=\left(\frac{2-a}{2a-1}\right)$ ....(3)
Also, from (2), we get
$3+{\log }_{5}7=b+3b{\log }_{5}7\Rightarrow (3-b)=(3b-1){\log }_{5}7\Rightarrow {\log }_{5}7=\left(\frac{3-b}{3b-1}\right)$ ....(4)
$\therefore$ From (3) and (4), we get
$\frac{2-a}{2a-1}=\frac{3-b}{3b-1}\Rightarrow (2-a)(3b-1)=(3-b)(2a-1)$
$\Rightarrow 6b-2-3ab+a=6a-3-2ab+b\Rightarrow 1=5a-5b+ab\Rightarrow \left(\frac{1-ab}{(a-b)}\right)=5$