Solve for x in each of the following :
Problem 1 :
4x = 64
Problem 2 :
9x - 1 = 27
Problem 3 :
8x = 1
Problem 4 :
Problem 5 :
9 ⋅ 2x = 72
Problem 6 :
Problem 7 :
Problem 8 :
Problem 9 :
3x - 1⋅ 92x + 1 = 243
Problem 10 :
9x = 7(3x) + 18
1. Answer :
4x = 64
4x = 43
x = 3
2. Answer :
9x - 1 = 27
(32)x - 1 = 33
32(x - 1) = 33
32x - 2 = 33
2x - 2 = 3
2x = 5
3. Answer :
8x = 1
8x = 80
x = 0
4. Answer :
5x - 2 = 5-2
x - 2 = -2
x = 0
5. Answer :
9 ⋅ 2x = 72
Divide both sides by 9.
2x = 8
2x = 23
x = 3
6. Answer :
(33)x - 1 = (3-1)1 - 2x
33(x - 1) = 3-1(1 - 2x)
33x - 3 = 3-1 + 2x
3x - 3 = -1 + 2x
x - 3 = -1
x = 2
7. Answer :
52(x + 2) = 5-3
52x + 4 = 5-3
2x + 4 = -3
2x = -7
8. Answer :
2x + 1 ⋅ (22)x = (2-1)x + 1
2x + 1 ⋅ 22x = 2-(x + 1)
2x + 1 + 2x = 2-x - 1
23x + 1 = 2-x - 1
3x + 1 = -x - 1
4x = -2
9. Answer :
3x - 1⋅ 92x + 1 = 243
3x - 1⋅ (32)2x + 1 = 35
3x - 1⋅ 32(2x + 1) = 35
3x - 1⋅ 34x + 2 = 35
3x - 1 + 4x + 2 = 35
35x + 1 = 35
5x + 1 = 5
5x = 4
10. Answer :
9x = 7(3x) + 18
(32)x = 7(3x) + 18
(3x)2 = 7(3x) + 18
Let y = 3x.
y2 = 7y + 18
y2 - 7y - 18 =0
y2 - 9y + 2y - 18 =0
y(y - 9) + 2(y - 9) =0
(y - 9)(y + 2) = 0
y - 9 = 0 or y + 2 = 0
y = 9 or y = -2
y - 9 = 0 y = 9 y = 32 3x = 32 x = 2 |
y + 2 = 0 y = -2 y = -2 3x = -2 |
In 3x, whatever real value (positive or negative or zero) we substitute for x, 3x can never be negative. So we can ignore the equation 3x = -2.
Therefore,
x = 2
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