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Question Number 178320 by Shrinava last updated on 15/Oct/22

Commented by mr W last updated on 15/Oct/22

$$onlyif\log 3means{\log }_{e}3,i.e.\ln 3$$

Commented by Acem last updated on 15/Oct/22

$$Goodsir$$

Answered by mr W last updated on 15/Oct/22

$$3=\frac{12}{4}>\frac{9}{4}$$$$\Rightarrow \sqrt{3}>\sqrt{\frac{9}{4}}=\frac{3}{2}$$$$$$$$2.5<e<3\left(proofseebelow\right)$$$$$$$$e^3>2\times 2\times 2.5=10>9$$$$\Rightarrow e^{\frac{3}{2}}>9^{\frac{1}{2}}=3$$$$$$$$e<3<e^{\frac{3}{2}}$$$$\ln e<\ln 3<\ln e^{\frac{3}{2}}$$$$\Rightarrow 1<\ln 3<\frac{3}{2}$$$$$$$$\frac{\pi \sqrt{3}}{9}<\frac{4\times 2}{9}=\frac{8}{9}<1<\ln 3$$$$\frac{\pi \sqrt{3}}{3}>\frac{3\sqrt{3}}{3}=\sqrt{3}>\frac{3}{2}>\ln 3$$$$\Rightarrow \frac{\pi \sqrt{3}}{9}<\ln 3<\frac{\pi \sqrt{3}}{3}✓$$$$$$$$===========$$$$prooffor\frac{5}{2}<e<3$$$$e=1+1+\frac{1}{2!}+\frac{1}{3!}+...>1+1+\frac{1}{2}=\frac{5}{2}$$$$e=1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}...$$$$=1+1+\frac{1}{2}+\frac{1}{2\times 3}+\frac{1}{2\times 3\times 4}+..$$$$<1+1+\frac{1}{2}+\frac{1}{2\times 2}+\frac{1}{2\times 2\times 2}+..$$$$=1+\frac{1}{1-\frac{1}{2}}=3$$

Commented by Tawa11 last updated on 15/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

Commented by Shrinava last updated on 15/Oct/22

$$\mathrm{cool}\mathrm{dear}\mathrm{professor}\mathrm{thank}\mathrm{you}$$

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