Find the greatest integer less than $3^\sqrt{3}$ without using a calculator and prove the answer is correct.

I'm puzzled on how to solve this problem, any help is appreciated. There was hints about turning the exponents into fractions and picking fractions between : $3^x < 3^\sqrt3 <3^y$

Then I simplified: $x< \sqrt3<y$

$x^2< 3<y^2$


So $x=\sqrt2$ and $y=\sqrt4=2$

$3^\sqrt2 < 3^\sqrt3 <3^2$

  • 1
    $\begingroup$ $\frac 32\lt\sqrt 3\lt\frac 74$ $\endgroup$
    – Andrei
    Commented Jul 15, 2020 at 13:26

2 Answers 2


Since $3 = \frac{48}{16} <\frac{49}{16}$, you have $\sqrt{3}<\frac{7}{4}.$ So you might try to take $y=\frac{7}{4}.$ It's easy to calculate $3^7 = 2187$ which is close to $2401 = 7^4.$ So $3^7<7^4$ and you have $3^{7/4}<7.$ So your answer is $6$ or less.

Try $x=5/3$ to get the lower bound.

  • $\begingroup$ Ingenious, why haven't I ever thought of it like that? Thanks. $\endgroup$
    – Anthony
    Commented Jul 15, 2020 at 15:12

Use the (easily verified) inequalities $\sqrt3\lt1.75=7/4$ and $\sqrt{7/4}\lt4/3$ to show that


and $\sqrt3\gt1.7$ to show


Now show $3^7=3^4\cdot3^3=81\cdot27\gt1600\gt1024=2^{10}$ to conclude $3^{7/10}\gt2$ and thus


Putting these together, we have $\lfloor3^\sqrt3\rfloor=6$.

Just to be thorough, let's verify the asserted inequalities:

$$\sqrt3\lt7/4\iff3\lt49/16\iff3\cdot16\lt49\\ \sqrt{7/4}\lt4/3\iff7/4\lt16/9\iff7\cdot9\lt4\cdot16\\ \sqrt3\gt1.7\iff3\gt1.7^2=2.89$$


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