Question

The solution set of In(5 – 7x) <1 is given by:
(-10,7)
All real numbers​

Answer 1

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{{\bf \: log(x) \: is \: defined \: when \: x > 0}}$

$\boxed{{\bf \: log_{y}(x) < z \implies \: x < {y}^{z} }}$

$\boxed{{\bf \: log(x) = log_{e}(x)}}$

$\large\underline{\bf{Solution-}}$

$\rm :\longmapsto\: log(5 - 7x) < 1$

can be rewritten as

$\rm :\longmapsto\: log_{e}(5 - 7x) < 1$

Case :- 1

$\rm :\longmapsto\:5 - 7x > 0$

$\rm :\longmapsto\: - 7x > - 5$

$\bf\implies \:x < \dfrac{5}{7} - - - (1)$

Case :- 2

$\rm :\longmapsto\: log_{e}(5 - 7x) < 1$

$\rm :\implies\:5 - 7x < {e}^{1}$

$\rm :\longmapsto\:5 - 7x < e$

$\rm :\longmapsto\: - 7x < e - 5$

$\rm :\implies\:x > \dfrac{5 - e}{7} - - - (2)$

So,

On combining equation (1) and (2), we get

$\bf\implies \:x \: \in \: \bigg(\dfrac{5 - e}{7} \: , \: \dfrac{5}{7} \bigg)$