Chapter 2: Exponents Of Real Numbers Exercise – 2.2
Question: 1
Assuming that x, y, z are positive real numbers, simplify each of the following
Solution:
= (243x10y5z10)1/5
= (243)1/5x10/5y5/5z10/5
= (35)1/5x2yz2
= 3x2yz2
Question: 2
Simplify
Solution:
= (4-1)
= 1/4
= [(25)−3]1/5
= (2−15)1/5
= 2−3
= 1/23 = 1/8
= [(343)−2]1/3
= (343)−2×1/3
= (73)−2/3
= (7−2)
=(1/72)
= (1/49)
(iv) (0.001)1/3
= (1/1000)1/3
= (1/103)1/3
= 721-20 × 525/2 - 21/2
= 71 × 54/2
= 71 × 52
= 7 × 25
= 175
Question: 3
Prove that
(ii) 93/2 − 3 × 50 − (1/81)−1/2
Solution:
= ((3 × 5−3)1/2 ÷ (3−1)1/3(5)1/2) × (3 × 56)1/6
= ((3)1/2(5−3)1/2 ÷ (3−1)1/3(5)1/2) × (3 × 56)1/6
= ((3)1/2(5)−3/2 ÷ (3)−1/3(5)1/2) × ((3)1/6 × (5)6/6)
= ((3)1/2 − (−1/3) × (5)−3/2−1/2) × ((3)1/6 × (5))
= ((3)5/6 × (5)−2) × ((3)1/6 × (5))
= ((3)5/6+1/6 × (5)−2+1)
= ((3)6/6 × (5)−1)
= ((3)1 × (5)−1)
= ((3) × (5)−1)
= ((3) × (1/5))
= (3/5)
(ii) 93/2 − 3 × 50 − (1/81)−1/2
= (32)3/2 − 3 − (1/92)−1/2
= 33 − 3 − (9)−2×−1/2
= 27 − 3 − 9
= 15
= 24 − 3 × 23×2/3 + 4/3
= 16 − 3 × 22 + 4/3
= 16 − 3 × 4 + 4/3
= 16 − 12 + 4/3
= (12 + 4)/3
= 16/3
= 2 × 1 × 5
= 10
= 1/2 + 1/(0.1)1 − (3)2
= 1/2 +1/(0.1) −9
= 1/2 + 10 − 9
= 1/2 + 1
= 3/2
= (5/4)2 + 5/4 + 5/4
= 25/16 + 10/4
= 25/16 + 40/16
= (26 + 40)/16
= 65/16
Question: 4
Show that
(v) (xa−b)a + b (xb−c)b + c(xc − a)c + a = 1
Solution:
Left hand side (LHS) = Right hand side (RHS)
Considering LHS,
Therefore, LHS = RHS
Hence proved
Left hand side (LHS) = Right hand side (RHS)
Considering LHS,
Therefore, LHS = RHS
Hence proved
(v) (xa−b)a + b (xb−c)b + c(xc − a)c + a = 1
(xa−b)a+b(xb−c)b+c(xc−a)c+a
Question: 5
If 2x = 3y = 12z, show that 1/z = 1/y + 2/x
Solution:
2x = 3y = (2 × 3 × 2)z
2x = 3y = (22 × 3)z
2x = 3y = (22z × 3z)
2x = 3y = 12z = k
2 = k1/x
3 = k1/y
12 = k1/z
12 = 2 × 3 × 2
12 = k1/z = k1/y × k1/x × k1/x
k1/z = k2/x + 1/y
1/z = 1/y + 2/x
Question: 6
If 2x = 3y = 6−z, show that 1/x + 1/y + 1/z = 0
Solution:
2x = 3y = 6−z
2x = k
2 = k1/x
3y = k
3 = k1/y
6−z = k
k = 1/6z
6 = k−1/z
2 × 3 = 6
k1/x × k1/y = k−1/z
1/x + 1/y = −1/z [by equating exponents]
1/x + 1/y + 1/z = 0
Question: 7
If ax = by = cz and b2 = ac, then show that
Solution:
Let ax = by = cz = k
a = k1/x, b = k1/y, c = k1/z
Now,
b2 = ac
(k1/y)2 = k1/x × k1/z
k2/y = k1/x + 1/z
2/y = 1/x + 1/z
Question: 8
If 3x = 5y = (75)z, Show that
Solution:
3x = k
3 = k1/x
5y = k
5 = k1/y
75z = k
75 = k1/z
31 × 52 = 751
k1/x × k2/y = k1/z
1/x + 2/y = 1/z
Question: 9
If (27)x = 9/3x, find x
Solution:
We have,
(27)x = 9/3x
(33)x = 9/3x
33x = 9/3x
33x = 32/3x
33x = 32−x
3x = 2 − x [On equating exponents]
3x + x = 2
4x = 2
x = 2/4
x = 1/2
Here the value of x is ½
Question: 10
Find the values of x in each of the following
(ii) (23)4 = (22)x
(iii) (3/5)x(5/3)2x = 125/27
(iv) 5x−2 × 32x−3 = 135
(v) 2x−7 × 5x−4 = 1250
(vii) 52x+3 = 1
Solution:
We have
= 4x = 4 [On equating exponent]
x = 1
Hence the value of x is 1
(ii) (23)4 = (22)x
We have
(23)4 = (22)x
= 23×4 = 22×x
12 = 2x
2x = 12 [On equating exponents]
x = 6
Hence the value of x is 6
(iii) (3/5)x(5/3)2x = 125/27
We have
(3/5)x(5/3)2x = 125/27
⇒ 52x−x/32x−x = 53/33
⇒ 5x/3x = 53/33
⇒ (5/3)x = (5/3)3
x = 3 [on equating exponents]
Hence the value of x is 3
(iv) 5x−2 × 32x−3 = 135
We have,
5x−2 × 32x−3 = 135
⇒ 5x−2 × 32x−3 = 5 × 27
⇒ 5x−2 × 32x−3 = 51 × 33
⇒ x − 2 = 1, 2x − 3 = 3 [On equating exponents]
⇒ x = 2 + 1, 2x = 3 + 3
⇒ x = 3, 2x = 6
⇒ x = 3
Hence the value of x is 3
(v) 2x−7 × 5x−4 = 1250
We have
2x−7 × 5x−4 = 1250
⇒ 2x−7 × 5x−4 = 2 × 625
⇒ 2x−7 × 5x−4 = 2 × 54
⇒ x − 7 = 1
⇒ x = 8, x − 4 = 4
⇒ x = 8
Hence the value of x is 8
4x + 1 = -15
4x = -15 - 1
4x = -16
x = (-16)/4
x = - 4
Hence the value of x is 4
(vii) 52x+3 = 1
52x+3 = 1 × 50
2x + 3 = 0 [By equating exponents]
2x = −3
x = −3/2
Hence the value of x is −3/2
√x = 2 [By equating exponents]
(√x)2 = (2)2
x = 4
Hence the value of x is 4
x + 1 = - 6
x = - 6 - 1
x = -7
Hence the value of x is 7
Question: 11
If x = 21/3 + 22/3, show that x3 − 6x = 6
Solution:
x3 − 6x = 6
x = 21/3 + 22/3
Putting cube on both the sides, we get
x3 = (21/3 + 22/3)3
As we know, (a + b)3 = a3 + b3 + 3ab(a + b)
x3 = (21/3)3 + (22/3)3 + 3(21/3)(22/3)(21/3 + 22/3)
x3 = (21/3)3 + (22/3)3 + 3(21/3+2/3)(x)
x3 = (21/3)3 + (22/3)3 + 3(2)(x)
x3 = 6 + 6x
x3 - 6x = 6
Hence proved
Question: 12
Determine (8x)x, if 9x+2 = 240 + 9x.
Solution:
9x+2 = 240 + 9x
9x .92 = 240 + 9x
Let 9x be y
81y = 240 + y
81y - y = 240
80y = 240
y = 3
Since, y = 3
Then,
9x = 3
32x = 3
Therefore, x = ½
(8x)x = (8 × 1/2)1/2
= (4)1/2
= 2
Therefore (8x)x = 2
Question: 13
If 3x+1 = 9x-2, find the value of 21+x
Solution:
3x+1 = 9x-2
3x+1 = 32x-4
x + 1 = 2x - 4
x = 5
Therefore the value of 21+x = 21+5 = 26 = 64
Question: 14
If 34x = (81)-1 and (10)1/y = 0.0001, find the value of 2-x+4y.
Solution:
34x = (81)-1 and (10)1/y = 0.0001
34x = (3)-4
x = -1
And, (10)1/y = 0.0001
(10)1/y = (10)−4
1/y = -4
y = 1/−4
To find the value of 2-x+4y, we need to substitute the value of x and y
2-x+4y = 21+4(1/−4) = 21-1 = 20 = 1
Question: 15
If 53x = 125 and 10y = 0.001. Find x and y.
Solution:
53x = 125 and 10y = 0.001
53x = 53
x = 1
Now,
10y = 0.001
10y = 10-3
y = -3
Therefore, the value of x = 1 and the value of y = – 3
Question: 16
Solve the following equations
(i) 3x+1 = 27 × 34
(iii) 3x−1 × 52y−3 = 225
(iv) 8x+1 = 16y+2 and (1/2)3+x = (1/4)3y
Solution:
(i) 3x+1 = 27 × 34
3x+1 = 33 × 34
3x+1 = 33+4
x + 1 = 3 + 4 [By equating exponents]
x + 1 = 7
x = 7 − 1
x = 6
4x = 3 (By equating exponents)
(iii) 3x−1 × 52y−3 = 225
3x−1 × 52y−3 = 32 × 52
x − 1 = 2 [By equating exponents]
x = 3
3x−1 × 52y−3 = 32 × 52
2y − 3 = 2 [By equating exponents]
2y = 5
y = 5/2
(iv) 8x+1 = 16y+2 and (1/2)3+x = (1/4)3y
(23)x+1 and (2−1)3+x = (2−2)3y
3x + 3 = 4y + 8 and − 3 − x = −6y
3x + 3 = 4y + 8 and 3 + x = 6y
3x + 3 = 4y + 8 and y = (3+x)/6
3x + 3 = 4y + 8... eq1
Substitute eq2 in eq1
3(3x + 3) = 6 + 2x + 24
9x + 9 = 30 + 2x
7x = 21
x = 21/7
x = 3
Putting value of x in eq2
y = 1
(v) 4x−1 × (0.5)3−2x = (1/8)x
22x−2 × (5/10)3−2x = (1/23)x
22x−2 × (1/2)3−2x = 2−3x
22x−2 × 2−3+2x = 2−3x
2x − 2 − 3 + 2x = −3x [By equating exponents]
4x + 3x = 5
7x = 5x = 5/7
(a/b)1/2 = (a/b)−(1−2x)1/2 = −1 + 2x [By equating exponents]
1/2 + 1 = 2x
2x = 3/2
x = ¾
Question: 17
If a and b are distinct positive primes such that
, find x and y
Solution:
(a6b−4)1/3 = axb2y
a6/3b−4/3 = axb2y
a2b−4/3 = axb2y
x = 2, 2y = −4/3
Question: 18
If a and b are different positive primes such that
(ii) (a + b)−1(a−1 + b−1) = axby, find x and y
Solution:
(a−1−2b2+4)7 ÷ (a3+2b−5−3) = axby
(a−3b6)7 ÷ (a5b−8) = axby
(a−21b42) ÷ (a5b−8) = axby
(a−21−5b42+8) = axby
(a−26b50) = axby
x = −26, y = 50
(ii) (a + b)−1(a−1 + b−1) = axby, find x and y
(a + b)−1(a−1 + b−1)
= 1/ab
= (ab)−1 = a−1b−1
By equating exponents
x = −1, y = −1
Therefore x + y + 2 = −1 − 1 + 2 = 0
Question: 19
If 2x × 3y × 5z = 2160, find x, y and z. Hence compute the value of 3x × 2−y × 5−z
Solution:
2x × 3y × 5z = 2160
2x × 3y × 5z = 24 × 33 × 51
x = 4, y = 3, z = 1
3x × 2−y × 5−z = 34 × 2−3 × 5−1
= 81/40
Question: 20
If 1176 = 2a × 3b × 7c, find the values of a, b and c.
Solution:
Hence compute the value of 2a × 3b × 7−c as a fraction
1176 = 2a × 3b × 7c
23 × 31 × 72 = 2a × 3b × 7c
a = 3, b = 1, c = 2
We have to find the value of 2a × 3b × 7−c
2a × 3b × 7−c = 23 × 31 × 7−2
= 24/49
Question: 21
Simplify
Solution:
(xa+b−c)a−b(xb+c−a)b−c(xc+a−b)c−a
Question: 22
Show that
Solution:
Hence, LHS = RHS
Question: 23
(i) If a = xm+nyl, b = xn+lym and c = xl+myn, prove that am−nbn−lcl−m = 1
(ii) If x = am+n, y = an+l and z = al+m, prove that xmynzl = xnylzm
Solution:
(i) If a = xm+nyl, b = xn+lym and c = xl+myn, prove that am−nbn−lcl−m = 1
(xm+nyl)m−n(xn+lym)n−l(xl+myn)l−m
= (x(m+n)(m−n)yl(m−n))(x(n+l)(n−l)ym(n−l))(x(l+m)(l−m)yn(l−m))
= x0y0 = 1
(ii) If x = am+n, y = an+l and z = al+m, prove that xmynzl = xnylzm
LHS = xmynzl
(am+n)m(an+l)n(al+m)l
=a(m+n)na(n+l)la(l+m)m
= xnylzm