# Compare an expression with zero

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

With the use of $$\;9 we get $$\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.$$
It's easy to take conservative estimates $$3^{-\frac{2}{3}}<\frac{1}{2}$$, $$\log_e9<\frac{5}{2}$$ and conclude that the expression is positive.
• And why is $\log_e 3$ less than $\frac{5}{2}$? – Mark Tiukov Nov 15 '18 at 12:22
• @МаркТюков: because $e\approx 2.7$, $e^2\approx 7.3$, $\sqrt{e} > 1.5$ – Vasya Nov 15 '18 at 12:41