Question

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

Answer 1

$a={\log }_{245}175=\frac{\log 175}{\log 245}=\frac{\log (25\times 7)}{\log (5\times 49)}=\frac{2\log 5+\log 7}{\log 5+2\log 7}$

$[\because {\log }_{b}a=\frac{\log a}{\log b},\log (ab)=\log a+\log b,\log a^m=m\log a]$

And $b={\log }_{1715}875=\frac{\log 875}{\log 1715}=\frac{\log (125\times 7)}{\log (5\times 343)}=\frac{3\log 5+\log 7}{\log 5+3\log 7}$

Consider, $c=\frac{1-ab}{a-b}=\frac{1-(\frac{2\log 5+\log 7}{\log 5+2\log 7}\times \frac{3\log 5+\log 7}{\log 5+3\log 7})}{\frac{2\log 5+\log 7}{\log 5+2\log 7}-\frac{3\log 5+\log 7}{\log 5+3\log 7}}$

$=\frac{(\log 5+2\log 7)(\log 5+3\log 7)-(2\log 5+\log 7)(3\log 5+\log 7)}{(2\log 5+\log 7)(\log 5+3\log 7)-(\log 5+2\log 7)(3\log 5+\log 7)}$

$\Rightarrow c=\frac{5[{\left(\log 7\right)}^2-{\left(\log 5\right)}^2]}{{\left(\log 7\right)}^2-{\left(\log 5\right)}^2}=5$

Ans: 5