Question Number 73523 by 01 last updated on 13/Nov/19
$$\mathrm{prove}\:\mathrm{that}\::\: \\ $$$$\:\mathrm{2}^{\pi} >\mathrm{8} \\ $$
Answered by MJS last updated on 13/Nov/19
$$\mathrm{3}<\pi\:\Rightarrow\:\pi=\mathrm{3}+{p};\:{p}>\mathrm{0} \\ $$$$\mathrm{2}^{\mathrm{3}+{p}} >\mathrm{2}^{\mathrm{3}} \\ $$$$\mathrm{2}^{\mathrm{3}} \mathrm{2}^{{p}} >\mathrm{2}^{\mathrm{3}} \\ $$$$\mathrm{2}^{{p}} >\mathrm{1} \\ $$$${p}\mathrm{ln}\:\mathrm{2}\:>\mathrm{0} \\ $$$${p}>\mathrm{0} \\ $$$$\mathrm{true} \\ $$