Question

Solve the following inequality:
$(0.5{)}^{{\log }_5{\log }_{0.3}(x-0.7)}<1$

Answer 1

$2^{-{\log }_5{\log }_{0.3}(x-0.7)}<1$ and $x-0.7>0x>0.7x\in (0.7,\infty )$

Taking ${\log }_2$ both sides

$-{\log }_5{\log }_{0.3}(x-0.7)<0{\log }_5{\log }_{0.3}(x-0.7)>0{\log }_{0.3}(x-0.7)>1\frac{\log (x-0.7)}{\log 0.3}>1\log \left(0.3\right)\log (x-0.7)>{\log }^2\left(0.3\right){\log }^2\left(0.3\right)-\log \left(0.3\right)\log (x-0.7)<0\log \left(0.3\right)\left[\log \right(0.3)-\log (x-0.7\left)\right]<0\log \left(0.3\right)-\log (x-0.7)>0\log \left(0.3\right)>\log (x-0.7)0.3>x-0.71>xx\in (-\infty ,1)$

Taking intersection, we get $x\in (0.7,1)$