Find the greatest integer less than $3^\sqrt{3}$ without using a calculator and prove the answer is correct. 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 $y=\sqrt4=2$
$3^\sqrt2 < 3^\sqrt3 <3^2$

Use the (easily verified) inequalities $\sqrt3\lt1.75=7/4$ and $\sqrt{7/4}\lt4/3$ to show that
$$3^\sqrt3\lt3^{1.75}=3\cdot3^{1/2}\cdot3^{1/4}\lt3\cdot{7\over4}\cdot{4\over3}=7$$
and $\sqrt3\gt1.7$ to show
$$3^\sqrt3\gt3^{1.7}=3\cdot3^{7/10}$$
Now show $3^7=3^4\cdot3^3=81\cdot27\gt1600\gt1024=2^{10}$ to conclude $3^{7/10}\gt2$ and thus
$$6=3\cdot2\lt3\cdot3^{7/10}=3^{1.7}\lt3^\sqrt3$$
Putting these together, we have $\lfloor3^\sqrt3\rfloor=6$.
Just to be thorough, let's verify the asserted inequalities:
$$\sqrt3\lt7/4\iff3\lt49/16\iff3\cdot16\lt49\\
\sqrt{7/4}\lt4/3\iff7/4\lt16/9\iff7\cdot9\lt4\cdot16\\
\sqrt3\gt1.7\iff3\gt1.7^2=2.89$$
