Help me arrange \(180^{4}\),\(90^{8}\),\(60^{12}\),\(40^{18}\),\(30^{24}\),\(15^{48}\),\(10^{72}\),\(5^{144}\),\(4^{180}\),\(3^{240}\),\(80^{9}\),\(72^{10}\) in \[\color{BLUE}{\huge{ascending }}\]order.

Take the \( 720 \) th root of all numbers. Then, all the terms come into the form of \( x^{1/x} \). Since it is a decreasing function for \( x > e \), if \( y > x > e \), then \( x^{1/x} > y^{1/y} \). Therefore, the order is \( 3^{1/3} > 4^{1/4}> ... > 180^{1/180} \). Raising each term to the \( 720 \)th power, \( 3^{240} > 4^{180} > ... > 180^4 \)

First take LCM of all the powers and then bring all the numbers in the power equal to LCM which is 720.
Then the arrangement becomes
3^240,4^180,5^144,10^72,15^48,
30^24,40^18,60^12,72^10,80^9,90^8,180^4

## Comments

Sort by:

TopNewestTake the \( 720 \) th root of all numbers. Then, all the terms come into the form of \( x^{1/x} \). Since it is a decreasing function for \( x > e \), if \( y > x > e \), then \( x^{1/x} > y^{1/y} \). Therefore, the order is \( 3^{1/3} > 4^{1/4}> ... > 180^{1/180} \). Raising each term to the \( 720 \)th power, \( 3^{240} > 4^{180} > ... > 180^4 \)

Log in to reply

First take LCM of all the powers and then bring all the numbers in the power equal to LCM which is 720. Then the arrangement becomes 3^240,4^180,5^144,10^72,15^48, 30^24,40^18,60^12,72^10,80^9,90^8,180^4

Log in to reply