If we have power raised to another power, we have to multiply the powers.
If we have numerical values inside the parenthesis,
Simplify the following without using calculator.
Example 1 :
(-32)2/5
Solution :
= (-32)2/5
Decomposing 32, we get
-32 = (-2) ⋅ (-2) ⋅ (-2) ⋅ (-2) ⋅ (-2)
-32 = (-2)5
(-32)2/5 = [(-2)5]2/5
Since we have power raised to another power, we have to multiply the powers.
= (-2)5 ⋅ 2/5
= (-2)2
(-32)2/5 = 4
Example 2 :
(-8)4/3
Solution :
= (-8)4/3
Decomposing -8, we get
-8 = (-2) ⋅ (-2) ⋅ (-2)
-8 = (-2)3
(-8)4/3 = [(-2)3]4/3
= (-2)3
= -8
Example 3 :
(8/27)4/3
Solution :
= (8/27)4/3
Decomposing 8 and 27, we get
8 = 2 ⋅ 2 ⋅ 2 8 = 23 |
27 = 3 ⋅ 3 ⋅ 3 27 = 33 |
(8/27)4/3 = [(2/3)3]4/3
= (2/3) 3 ⋅ (4/3)
= (2/3)4
= 16/81
Example 4 :
(125)1/3
Solution :
= (125)1/3
Decomposing 125, we get
125 = 5 ⋅ 5 ⋅ 5 ==> 53
= (53)1/3
= 53 ⋅ (1/3)
= 5
Example 5 :
(8x15)-1/3
Solution :
= (8x15)-1/3
Decomposing 8, we get
8 = 2 ⋅ 2 ⋅ 2 ==> 23
= (23 (x5)3)-1/3
= (2x5)3 ⋅ (-1/3)
= (2x5)-1
=1/(2x5)
Example 6 :
(h6p9/1000m3)-2/3
Solution :
Example 7 :
(64)1/6
Solution :
= (64)1/6
Decomposing 64, we get
64 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2
64 = 26
(64)1/6 = (26)1/6
= 2 (6 ⋅ 1/6)
= 2
Example 8 :
(256)3/4
Solution :
= (256)3/4
Decomposing 256, we get
256 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2
256 = 44
(256)3/4 = (44)3/4
= 43
= 64
Example 8 :
evaluate the expression without using a calculator.
a) 641/6 b) 81/3 c) 253/2 d) 813/4 |
e) (−243)1/5 f) (−64)4/3 g) 8−2/3 h) 16−7/4 |
Solution :
a) 641/6
Step 1 :
Writing 64 in expanded form.
64 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2
Step 2 :
Writing the expanded form in exponential form, we get
= 26
Step 3 :
Replacing the exponential form in the question and multiplying the power raised by another power.
641/6 = (26)1/6
= 26 x (1/6)
= 2
b) 81/3
Step 1 :
Writing 8 in expanded form.
8 = 2 ⋅ 2 ⋅ 2
Step 2 :
Writing the expanded form in exponential form, we get
= 23
81/3 = (23)1/3
= 23x(1/3)
= 2
c) 253/2
25 = 5 ⋅ 5
= 52
253/2 = (52)3/2
= 52 x (3/2)
= 53
= 125
d) 813/4
Step 1 :
Writing 81 in expanded form.
81 = 9 ⋅ 9
Step 2 :
Writing the expanded form in exponential form, we get
= 34
813/4 = (92)3/4
= 92 x (3/2)
= 93
= 729
e) (−243)1/5
Step 1 :
Writing 243 in expanded form.
243 = 3 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 3
Step 2 :
Writing the expanded form in exponential form, we get
= 35
(−243)1/5 = ((-3)5)1/5
= (-3)5 x (1/5)
= -3
f) (−64)4/3
Step 1 :
Writing 64 in expanded form.
64 = 4 ⋅ 4 ⋅ 4
Step 2 :
Writing the expanded form in exponential form, we get
= 43
(−64)4/3 = ((-4)3)4/3
= (-4)3 x (4/3)
= (-4)4
We have negative base and even number as exponent, then the negative sign can be changes as positive.
= 256
g) 8−2/3
8 = 2 ⋅ 2 ⋅ 2 ==> 83
8−2/3 = (83)−2/3
= 8−2
= 1/82
= 1/64
h) 16−7/4
16 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ==> 24
16−7/4 = (24)−7/4
= 2−7
= 1/27
= 1/128
Example 9 :
a) 3(1111/4) + 9(1111/4)
b) 13(83/4) − 4(83/4)
c) 27√6 + 7√150
d) ∜(1296/25)
Solution :
a) 3(1111/4) + 9(1111/4)
Since these two are like terms, we can add them.
= 12(1111/4)
It cannot be simplified further.
b) 13(83/4) − 4(83/4)
= 9(83/4)
= 9((23)3/4)
= 9(29/4)
c) 27√6 + 7√150
√150 = √(5 ⋅ 5 ⋅ 3 ⋅ 2)
= 5√6
27√6 + 7√150 = 27√6 + 5√6
= 33√6
d) ∜(1296/625)
1296 = 6 ⋅ 6 ⋅ 6 ⋅ 6 ==> 64
625 = 5 ⋅ 5 ⋅ 5 ⋅ 5 ==> 54
∜(1296/625) = ∜(64/54)
= ∜(6/5)4
= 6/5
Example 10 :
Perform the indicated operation. Assume all variables are positive.
a) 12 ∛y + 9 ∛y
b) 11 √2z − 5√2z
c) 3x7/2− 5x7/2
d) 7 ∛m7 + 3m7/3
Solution :
a) 12 ∛y + 9 ∛y
= 21∛y
b) 11 √2z − 5√2z
= 6√2z
c) 3x7/2− 5x7/2
= -2 x7/2
d) 7 ∛m7 + 3m7/3
= 7 ∛m7 + 3m7x(1/3)
= 7 ∛m7 + 3 ∛m7
= 10∛m7
Example 11 :
Perform the indicated operation. Assume all variables are positive.
a) 1254/3
b) 324/5
c) 6253/4
d) 493/2
Solution :
a) 1254/3
= (53)(4/3)
= 53 x (4/3)
= 54
= 625
b) 324/5
= (25)(4/5)
= 25 x (4/5)
= 24
= 16
c) 6253/4
= (54)(3/4)
= 54 x (3/4)
= 53
= 125
d) 493/2
= (72)(3/2)
= 72 x (3/2)
= 73
= 343
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM