2.1.3 Powers of 2, 3, 5

Memorizing powers of certain integers like 2, 3, 5, etc… can be beneficial in solving a variety of problems ranging from approximation problems to logarithm problems. In some instances, powers of integers can be calculated based on other means than memorization. For example, 74=(72)2=492=24017^4 = (7^2)^2 = 49^2 = 2401. However, the following powers should be memorized for quick calculation:

Powers of 2Powers of 3Powers of 5
23=82^3 = 833=273^3 = 2753=1255^3 = 125
24=162^4 = 1634=813^4 = 8154=6255^4 = 625
25=322^5 = 3235=2433^5 = 24355=31255^5 = 3125
26=642^6 = 6436=7293^6 = 729
27=1282^7 = 12837=21873^7 = 2187
28=2562^8 = 256
29=5122^9 = 512
210=10242^{10} = 1024

Problem Set 2.1.3

53+33+235^3 + 3^3 + 2^3
23−33−432^3 - 3^3 - 4^3
(64−36)5(\sqrt{64} - \sqrt{36})^5
5x=125,x55^x = 125, x^5
43−534^3 - 5^3
2x+1=32,x−12^{x+1} = 32, x - 1
23+33+532^3 + 3^3 + 5^3
53−335^3 - 3^3
1253×512\sqrt[3]{125} \times 512
23+33+43−532^3 + 3^3 + 4^3 - 5^3
x3=64, so 3xx^3 = 64, \text{ so } 3x
45×554^5 \times 5^5
27227^2
If x5=−32, then 5x\text{If } x^5 = -32, \text{ then } 5x
25×532^5 \times 5^3
84×548^4 \times 5^4
55+44+33+22+115^5 + 4^4 + 3^3 + 2^2 + 1^1 *
26×542^6 \times 5^4
5x−1=3125, then x+15^{x-1} = 3125, \text{ then } x + 1
23−33−532^3 - 3^3 - 5^3
(34)2×3×53(\frac{3}{4})^2 \times 3 \times 5^3
63+43+236^3 + 4^3 + 2^3
34+43=5⋅x, then x3^4 + 4^3 = 5 \cdot x, \text{ then } x
51+42+33+24+155^1 + 4^2 + 3^3 + 2^4 + 1^5 *
9x=243, then x9^x = 243, \text{ then } x
83×538^3 \times 5^3
23×83×532^3 \times 8^3 \times 5^3
25×34×522^5 \times 3^4 \times 5^2
24×72×532^4 \times 7^2 \times 5^3
42×52×624^2 \times 5^2 \times 6^2
25×33×522^5 \times 3^3 \times 5^2
23×34×552^3 \times 3^4 \times 5^5
(33−23+13)×53(3^3 - 2^3 + 1^3) \times 5^3
25×34×522^5 \times 3^4 \times 5^2
25×34×552^5 \times 3^4 \times 5^5
23×32×42×532^3 \times 3^2 \times 4^2 \times 5^3