NomanSaleem wrote:

Kindly help me out with this question,

Q- Arrange 2^120, 3^72, 17^13 in ascending order.

I know we have to make either powers same or the bases but i am failing to do so.

Regards,

Hi

NomanSaleem,

Try to convert all the term in the power of 2s. We can't convert exactly each term in the power of 2s, so we'll find the range.

\(3 ^{72} = (3^{3})^{24} = 27^{24} < 32^{24} = 2^{120}\).

Hence, we have following:

\(2^{72} < 3^{72} < 2^{120}\)

Similarly, \(17^{13} < 32^{13} = 2^{65}\).

Ascending order: \(17^{13}, 3^{72}, 2^{120}\)

Hope this helps.