arrange the following in ascending order:
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Numbers are to be arranged in ascending order.
![\sqrt[8]{90},\ \ \ \sqrt[4]{10},\ \ \ \sqrt6](https://tex.z-dn.net/?f=%5Csqrt%5B8%5D%7B90%7D%2C%5C+%5C+%5C+++%5Csqrt%5B4%5D%7B10%7D%2C%5C+%5C+%5C+%5Csqrt6 "\sqrt[8]{90},\ \ \ \sqrt[4]{10},\ \ \ \sqrt6")
We have 8th root of 90. So we take 6 raised to 4 and 10 raised to 2, to start with along with 90.
we take an initial set :
![we\ know\ 81<90<100<36*36\\\\\sqrt{81}\ \textless \ \sqrt{90}\ \textless \ \sqrt{100}\ \textless \ \sqrt{36*36}\\\\So\ \ 9\ \textless \ \sqrt{90}\ \textless \ 10\ \textless \ 36\\\\3\ \textless \ \sqrt[4]{90}\ \textless \ \sqrt{10}\ \textless \ 6\\\\\sqrt3\ \textless \ \sqrt[8]{90}\ \textless \ \sqrt[4]{10}\ \textless \ \sqrt6](https://tex.z-dn.net/?f=we%5C+know%5C+81%26lt%3B90%26lt%3B100%26lt%3B36%2A36%5C%5C%5C%5C%5Csqrt%7B81%7D%5C+%5Ctextless+%5C+%5Csqrt%7B90%7D%5C+%5Ctextless+%5C+%5Csqrt%7B100%7D%5C+%5Ctextless+%5C+%5Csqrt%7B36%2A36%7D%5C%5C%5C%5CSo%5C+%5C+9%5C+%5Ctextless+%5C+%5Csqrt%7B90%7D%5C+%5Ctextless+%5C+10%5C+%5Ctextless+%5C+36%5C%5C%5C%5C3%5C+%5Ctextless+%5C+%5Csqrt%5B4%5D%7B90%7D%5C+%5Ctextless+%5C+%5Csqrt%7B10%7D%5C+%5Ctextless+%5C+6%5C%5C%5C%5C%5Csqrt3%5C+%5Ctextless+%5C+%5Csqrt%5B8%5D%7B90%7D%5C+%5Ctextless+%5C+%5Csqrt%5B4%5D%7B10%7D%5C+%5Ctextless+%5C+%5Csqrt6 "we\ know\ 81<90<100<36*36\\\\\sqrt{81}\ \textless \ \sqrt{90}\ \textless \ \sqrt{100}\ \textless \ \sqrt{36*36}\\\\So\ \ 9\ \textless \ \sqrt{90}\ \textless \ 10\ \textless \ 36\\\\3\ \textless \ \sqrt[4]{90}\ \textless \ \sqrt{10}\ \textless \ 6\\\\\sqrt3\ \textless \ \sqrt[8]{90}\ \textless \ \sqrt[4]{10}\ \textless \ \sqrt6")
The ascending order is clear from the above relation.
We have 8th root of 90. So we take 6 raised to 4 and 10 raised to 2, to start with along with 90.
we take an initial set :
The ascending order is clear from the above relation.
kvnmurty:
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you could also solve the exercise by squaring the given quantities 3 times. Then you will get 90, 100 and 6^4. Then we know which is greater.
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