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Question Number 73523 by 01 last updated on 13/Nov/19

prove that :    2^π >8

$$\mathrm{prove}\:\mathrm{that}\::\: \\ $$ $$\:\mathrm{2}^{\pi} >\mathrm{8} \\ $$

Answered by MJS last updated on 13/Nov/19

3<π ⇒ π=3+p; p>0  2^(3+p) >2^3   2^3 2^p >2^3   2^p >1  pln 2 >0  p>0  true

$$\mathrm{3}<\pi\:\Rightarrow\:\pi=\mathrm{3}+{p};\:{p}>\mathrm{0} \\ $$ $$\mathrm{2}^{\mathrm{3}+{p}} >\mathrm{2}^{\mathrm{3}} \\ $$ $$\mathrm{2}^{\mathrm{3}} \mathrm{2}^{{p}} >\mathrm{2}^{\mathrm{3}} \\ $$ $$\mathrm{2}^{{p}} >\mathrm{1} \\ $$ $${p}\mathrm{ln}\:\mathrm{2}\:>\mathrm{0} \\ $$ $${p}>\mathrm{0} \\ $$ $$\mathrm{true} \\ $$

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