What is the decreasing order from left to right of the following numbe

What is the decreasing order from left to right of the following numbers?

\(\sqrt[4]{3}, \sqrt[5]{6},\sqrt[10]{12},\sqrt[15]{45}\)

Let us get all the terms without roots, so raise each to the power of LCM(4,5,10,15) = 60
\(\sqrt[4]{3}, \sqrt[5]{6},\sqrt[10]{12},\sqrt[15]{45}\)=\(3^{15},6^{12},12^{6},45^4\)
(I) let us check between \(3^{15},6^{12}\)
\(3^{15}=3^{12}*3^3\) and \(6^{12}=3^{12}*2^{12}=3^{12}*(2^4)^3=3^{12}*16^3\).....\(3^{15}<6^{12}...\sqrt[4]{3}<\sqrt[5]{6}.\)..
Choices A and E can be eliminated
Similarly we can test other pairs..

Another way to test \(3^{15},6^{12},12^{6},45^4\) is to try to get them in same power
\(3^{15}=(3^5)^3=243^3,6^{12}=(6^4)^3=(1296)^3,12^{6}=(12^2)^3=(144)^3,45^4\)...
So, descending order \( \sqrt[5]{6},\sqrt[4]{3},\sqrt[10]{12},\), and we have to check where \(\sqrt[15]{45}\) fits in...
\(6^{12}=(6^3)^4=216^4,45^4\)......6^{12} >45^4
\(12^{6}=(12^3)^2=1728^2,45^4=(45^2)^2=(2025)^2.....45^4>12^6\)
\(3^{15}=3^8*3^7=3^8*3*(3^3)^2=3^8*3*27^2,45^4=3^8*5^4=3^8*25^2......3^{15}>45^4\)


(D) \(\sqrt[5]{6}, \sqrt[4]{3}, \sqrt[15]{45}, \sqrt[10]{12}\)