Please solve the questions
The answer to Question 2 is as below
5. Find x, if:
[∛(⅔)]x-1 = 27/8
SOLUTION
[(?)1/3]x-1 = (3 × 3 × 3)/(2 × 2 × 2)
(?)(x-1)/3 = (3/2)3
(?)(x-1)/3 = (2/3)-3
Now, if the bases are equal, then the powers must be equal
So, on comparing the exponents, we get
(x – 1)/3 = -3
x – 1 = -9
x = -9 + 1
x = -8
6. Solve for x:
(49)x + 4 = 72 × (343)x + 1
SOLUTION
We have, (49)x + 4 = 72 × (343)x + 1
(7 x 7)x + 4 = 72 × (7 x 7 x 7)x + 1
(72)x + 4 = 72 × (73)x + 1
(7)2x + 8 = (7)3x + 3 + 2
(7)2x + 8 = (7)3x + 5
Now, if the bases are equal, then the powers must be equal
On comparing the exponents, we get
2x + 8 = 3x + 5
3x – 2x = 8 – 5
x = 3
4. Show that:
(am/a-n)m-n × (an/a-l)n-l × (al/a-m)l-m = 1
SOLUTION
Taking the L.H.S., we have
(am/a-n)m-n × (an/a-l)n-l × (al/a-m)l-m
= (am × an)m-n × (an ×al)n-l × (al × am)l-m
= (am+n) m-n × (an+l)n-l × (al+m)l-m
= a0
= 1
1. Evaluate:
(i) √¼ + (0.01)-1/2 – (27)2/3
SOLUTION
v¼ + (0.01)-1/2 – (27)2/3 = v(½ x ½) + (0.1 x 0.1)-1/2 – (3 x 3 x 3)2/3
= v(½)2 + (0.12) -1/2 – (33)2/3
= ½ + (0.1)2 x -1/2 – (3)3 x 2/3
= ½ + (0.1)-1 – (3)2
= ½ + 1/0.1 – 32
= ½ + 1/(1/10) – 9
= ½ + 10 – 9
= ½ + 1
= 3/2
3. If 2160 = 2a. 3b. 5c, find a, b and c. Hence, calculate the value of 3a x 2-b x 5-c.
SOLUTION
We have,
2160 = 2a x 3b x 5c
(2 x 2 x 2 x 2) x (3 x 3 x 3) x 5 = 2a x 3b x 5c
24 x 33 x 51 = 2a x 3b x 5c
? 2a x 3b x 5c = 24 x 33 x 51
Comparing the exponents of 2, 3 and 5 on both sides, we get
a = 4, b = 3 and c = 1
Hence, the value
3a x 2-b x 5-c = 34 x 2-3 x 5-1
= (3 x 3 x 3 x 3) x (½ x ½ x ½) x 1/5
= 81 x 1/8 x 1/5
= 81/40
2. Simplify each of the following and express with positive index:
(i) (3-4/2-8)1/4
(ii) (27-3/9-3)1/5
SOLUTION
(i) (3-4/2-8)1/4 = (28/34)1/4
= (28)1/4/(34)1/4
= (2)8/4/(3)4/4
= 22/3
= 4/3
(ii) (27-3/9-3)1/5 = (93/273)1/5
= [(3 x 3)3/(3 x 3 x 3)3]1/5
= [(32)3/(33)3]1/5
= [(3)2 x 3/(3)3 x 3]1/5
= [(3)6/(3)9]1/5
= [(3)6-9]1/5
= (3)-3 x 1/5
= (3)-3/5
= 1/33/5
A stream of water flowing horizontally with a speed of 15 m/s pushes out of a tube of cross-sectional area 10-2 m2 and hits at a vertical wall nearby. What is the force exefrted on the wall by the impact of water, assuming that it does not rebound?
A helicopter of mass 1000 kg rises with a vertical acceleration of 15 ms-2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of
(a) force on the floor by the crew and passengers,
(b) the action of the rotor of the helicopter on surrounding air
(c) force on the helicopter due to the surrounding air
Mass of helicopter = 1000 kg
Crew and passengers weight = 300 kg
Vertical acceleration, a = 15 ms-2 and g = 10 ms-2
The total mass of the system, mi = 1000 + 300 = 1300 Kg
(a)Force on the floor of the helicopter by the crew and passengers
R – mg = ma
= m (g+a)
= m (g + a) = 300 (10 + 15) N = 7500 N
(b)Action of the rotor of the helicopter on surrounding air is due to the mass of the helicopter and the passengers.
R’ – mig = mia
R’ = mi (g+a)
= 1300 x (10 + 15) = 32500 N
This force acts vertically downwards
(c)Force on the helicopter due to the surrounding air is the reaction of the force applied by the rotor on the air. As action and reaction are equal and opposite, therefore, the force of reaction, F = 32500 N. This force acts vertically upwards
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