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: $x< \sqrt3<y$
$x^2< 3<y^2$
$\sqrt2^2<3<\sqrt4^2$
So $x=\sqrt2$ and $y=\sqrt4=2$
$3^\sqrt2 < 3^\sqrt3 <3^2$
Best Answer
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.