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Anonymous users2024-02-06Solution: a=3 to the 55th power = (3 to the fifth power) to the 11th power.
b = 4 to the power of 4 = (4 to the power of 4) to the power of 11.
c = 5 to the 33rd power = (5 to the 3rd power) to the 11th power.
So a = 243 to the 11th power.
b = 256 to the 11th power.
c = 125 to the 11th power.
So b a c. Please click "For Answer".
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Anonymous users2024-02-05This problem will be complicated if it is calculated, and the solution is as follows:
Notice that there is a common factor of 11, so there is.
It is clear: 4 44> 3 55> 5 33
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Anonymous users2024-02-04Easy way to calculate:
A = 3 to the 55th power = 3 to the 5th power and then 11 to the 11th power = 243 to the 11th power.
b = 4 to the power of 4 = 4 to the power of 4 and then to the power of 11 = 256 to the power of 11.
c = 5 to the 33rd power = 5 to the 3rd power and then 11 to the 11th power = 125 to the 11th power.
Therefore: b>a>c
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Anonymous users2024-02-03A 3 to the power of 55.
3 times 5 to the 11th power.
3 to the 5th power plus 3 to the 11th power.
243 to the 11th power.
b 4 to the power of 44.
4 times 4 times 11 power.
4 to the power of 4 plus 3 to the power of 11.
256 to the 11th power.
C 5 to the 33rd power.
5 times 3 times 11 power.
5 to the 3rd power plus 5 to the 11th power.
125 to the 11th power.
b>a>c
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Anonymous users2024-02-02It's convenient, and you don't have to do any homework.
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Anonymous users2024-02-01Fully endorse the first reply.
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Anonymous users2024-01-31Also to the power of 11.
4 to the 44th power, 3 to the 55th power, 5 to the 33rd power, 4 to the 44th power.
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Anonymous users2024-01-30If you put the number of 11 squares, you get the 11th power of the 5th power of a=2, the 11th power of the 4th power of b=3, and the 11th power of the 3rd power of c=4.
Then we get a 5th power of a=2, a 4th power of b=3, and a 3rd power of c=4.
This gives a=32 b=81 c=56
In the end, b》c》a
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Anonymous users2024-01-29a=（3）55=（（3)5)11=(243)11;b=(4)44=((4)4)11=(256)11;c=(5)33=((5)3)11=(125)11;So, b>a>c, because the upper corner mark is really hard to play, are they all students, should be able to understand ah ......This kind of problem should be seen in the common denominator, and it should be thought of after seeing 33, 44, 55.
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Anonymous users2024-01-28A = 2 to the power of 55 = (2 to the power of 5) to the power of 11 B = 3 to the power of 44 = (3 to the power of 4) to the power of 11 C = 4 to the power of 33 = (4 to the power of 3) to the power of 11 B>c>a
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Anonymous users2024-01-27a=2 55=2 (5*11)=(2 5) 11=32 11b=3 44=3 (4*11)=(3 4) 11=81 11c=4 33=4 (3*11)=(4 3) 11=64 11When the index is the same, the larger the base, the larger the value.
So a c b
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