2007-08-24 06:49:31 · 12 answers · asked by Pranil 7 in Science & Mathematics ➔ Mathematics
we know 2^3 = 8
3^2 -= 9
so 2^3 < 3^2
raise to the power 14
2^42 < 3^28 so 2^40 as < 2^42 is < 3^28
so 3^28 is greater
2007-08-25 01:34:22 · answer #1 · answered by Mein Hoon Na 7 · 2⤊ 0⤋
It's 3^28
2007-08-25 09:38:40 · answer #2 · answered by GB 2 · 0⤊ 0⤋
Taking log
log 2 ^ 40 = 40 log 2 = 40 x 0.3010 = 12.04
log 3 ^ 28 = 28 log 3 = 28 x 0.4771 = 13.3588
12.04 < 13.3588
Hence, 2 ^ 40 < 3 ^ 28
3 ^ 28 is the greater number
2007-08-25 06:15:39 · answer #3 · answered by pereira a 3 · 0⤊ 0⤋
A quick way to do this is to take the fourth root of both numbers and compare the reduced numbers.
2^10 and 3^7
2^10 = 1024
3^7 = 3(9^3) = 3*729, which is obviously larger than 1024.
since 3^7 > 2^10, 3^28 > 2^40. So 3^28 wins
2007-08-24 14:02:54 · answer #4 · answered by Derek C 3 · 2⤊ 0⤋
3^28
2007-08-26 01:44:02 · answer #5 · answered by vr n 2 · 0⤊ 0⤋
2^40 = 1099511627776
3^28 = 22876792454961
2007-08-24 13:58:51 · answer #6 · answered by Barkley Hound 7 · 0⤊ 0⤋
3^28
2007-08-24 13:54:06 · answer #7 · answered by Michael C 3 · 0⤊ 0⤋
Let x=2^40 and
y=3^28
take log of both sides
log x=40 log2=40*0,30103
log y =28 log3=28*0,47712
do the multiplications and compare log x and log y
2007-08-24 14:08:04 · answer #8 · answered by iyiogrenci 6 · 0⤊ 0⤋
2^40 / 3^28
= (2/3)^28 * 2^12
= (4/9)^14 * 2^12
= (4/9)^12 * (4/9)^2 * 2^12
= (8/9)^12 * (4/9)^2 <1
* because 8/9 and 4/9 are less than 1, so (8/9)^12 and (4/9)^2 are less than 1, too.
Therefore,
2^40 / 3^28 < 1
2^40 < 3^28.
2007-08-24 15:36:14 · answer #9 · answered by EWANG 2 · 0⤊ 0⤋
2^40 = 1 099 511 627 776
3^28 = 2.28767925 Ã 10^13
2007-08-24 13:55:15 · answer #10 · answered by Anonymous · 0⤊ 0⤋