Question Number 182702 by CrispyXYZ last updated on 13/Dec/22 | ||
$$\mathrm{log}_{\mathrm{2}} \left({x}+\mathrm{1}\right)−{x}>\mathrm{0} \\ $$ | ||
Commented bymr W last updated on 13/Dec/22 | ||
$${at}\:{x}=\mathrm{0}\:{and}\:{x}=\mathrm{1}:\:\mathrm{log}_{\mathrm{2}} \:\left({x}+\mathrm{1}\right)={x} \\ $$ $$\Rightarrow−\mathrm{1}<{x}<\mathrm{0}\:\vee\:\mathrm{1}<{x}<+\infty \\ $$ | ||