Compare an expression with zero

Question

I need to compare $1-\frac{2}{3}\cdot3^{-\frac{2}{3}}\cdot \log_e9$ and $0$ without any computer

You just need to see if it's positive or not. After some fiddling, you'll see that this is equivalent to seeing if $\displaystyle\log_e9 < \frac{3^\frac{5}{3}}{2}$. — user3482749, Nov 15 '18 at 12:02

score 0 · Answer 1 · answered Nov 15 '18 at 12:12

0

It's easy to take conservative estimates $3^{-\frac{2}{3}}<\frac{1}{2}$, $\log_e9<\frac{5}{2}$ and conclude that the expression is positive.

answered Nov 15 '18 at 12:12

Vasili

10,690

And why is $\log_e 3$ less than $\frac{5}{2}$? – Mark Tiukov Nov 15 '18 at 12:22
@МаркТюков: because $e\approx 2.7$, $e^2\approx 7.3$, $\sqrt{e} > 1.5$ – Vasili Nov 15 '18 at 12:41

user376343 · Accepted Answer · 2018-11-15T12:20:51.840

0

With the use of $\;9<e^3$ we get $$\frac{2}{3}\cdot3^{-\frac{2}{3}}\cdot \ln 9<\frac{2}{3}\cdot3^{-\frac{2}{3}}\cdot \ln e^3=\frac{2}{9^{1\over 3}}<\frac{2}{8^{1\over 3}}=1,$$ thus $$1-\frac{2}{3}\cdot3^{-\frac{2}{3}}\cdot \ln 9>0.$$

edited Nov 15 '18 at 12:20

answered Nov 15 '18 at 12:15

user376343

8,311

Compare an expression with zero

2 Answers2