Arrange irrationals in ascending order:$ 2^{\sqrt{\frac{5}{3}}},3^{\sqrt{\frac{3}{5}}},5^{\sqrt{\frac{4}{15}}},29^{\frac{1}{\sqrt{15}}} $ How to arrange following irrational numbers in ascending order:
$$ 2^{\sqrt{\frac{5}{3}}}, 3^{\sqrt{\frac{3}{5}}}, 5^{\sqrt{\frac{4}{15}}}, 29^{\frac{1}{\sqrt{15}}} $$
My try:
Using a calculator I computed the values
$$
2^{\sqrt{\frac{5}{3}}}\approx 2.445, \quad
3^{\sqrt{\frac{3}{5}}}\approx 2.342,\quad
5^{\sqrt{\frac{4}{15}}}\approx 2.296, \quad
29^{\frac{1}{\sqrt{15}}}\approx 2.385.
$$
By checking these values I easily arranged them in ascending order
$$ 5^{\sqrt{\frac{4}{15}}}<3^{\sqrt{\frac{3}{5}}}<
29^{\frac{1}{\sqrt{15}}}<2^{\sqrt{\frac{5}{3}}} $$
This is my final answer I found by calculator.

My question is: Can I arrange these irrational numbers without computing the values because I am not allowed to use calculator.

I have never seen the numbers with irrational powers.

Equate the powers by taking LCM of denominators and compare the base-numbers as follows
$$ 2^{\sqrt{\dfrac{5}{3}}}, 3^{\sqrt{\dfrac{3}{5}}}, 5^{\sqrt{\dfrac{4}{15}}}, 29^{\dfrac{1}{\sqrt{15}}} $$
$$ 2^{\dfrac{5}{\sqrt{15}}}, 3^{\dfrac{3}{\sqrt{15}}}, 5^{\dfrac{2}{\sqrt{15}}}, 29^{\dfrac{1}{\sqrt{15}}} $$
$$ (2^5)^{\dfrac{1}{\sqrt{15}}}, (3^3)^{\dfrac{1}{\sqrt{15}}}, (5^2)^{\dfrac{1}{\sqrt{15}}}, 29^{\dfrac{1}{\sqrt{15}}} $$
$$(32)^{\dfrac{1}{\sqrt{15}}}, (27)^{\dfrac{1}{\sqrt{15}}}, (25)^{\dfrac{1}{\sqrt{15}}}, 29^{\dfrac{1}{\sqrt{15}}} $$
Now, arrange above base numbers in increasing order as follows $$25<27<29<32$$
$$(25)^{\frac{1}{\sqrt{15}}}<(27)^{\frac{1}{\sqrt{15}}}<(29)^{\frac{1}{\sqrt{15}}}<(32)^{\frac{1}{\sqrt{15}}}$$
$$ \therefore 5^{\sqrt{\dfrac{4}{15}}}<3^{\sqrt{\dfrac{3}{5}}}<
29^{\dfrac{1}{\sqrt{15}}}<2^{\sqrt{\dfrac{5}{3}}} $$
