ICSE Class 9 Maths Selina Solution Chapter 7:
The ICSE Class 9 Maths Selina Solution Chapter 7 is useful for students because it helps them to get high marks in the exam. Selina Solutions contains a detailed step-by-step explanation of all the problems in Chapter 7 Indices [Exponents] of Selina's Class 9 Notebook. To score well in the ICSE Class 9 Maths exam, it is recommended to solve the questions and for your betterment, we have provided the ICSE Class 9 Maths Selina Solutions Chapter 7 Indices (Exponents).
ICSE Class 9 Maths Selina Solutions Chapter 7
In mathematics, the index is called the power or exponent to which the number is raised. For example, in the expression 3², 2 is the index of 3. The plural of index is indexes.
Simply put, the index determines how many times any number is multiplied by itself. Example: In the expression 2³ 2 is the base and 3 is the index. We read this term "2 to the power of 3" and it can be solved as : 2 × 2 × 2 = 8.
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ICSE Class 9 Maths Selina Solutions Chapter 7 PDF
We provide you with detailed answers and solutions for Selina's solutions for ICSE Class 9 Maths Chapter 7- Indices (Exponents). Subject experts have prepared these questions as per the syllabus prescribed by CISCE for ICSE. In our article we have provided ICSE Class 9 Mathematics Selina Solution Chapter 7 PDF. Here ICSE Class 9 Mathematics Selina Solutions Chapter 7 PDF file can be downloaded and viewed online as we have included direct download link below for easy and convenient access for candidates. Students can use and download these Selina solutions for offline practice for free.
ICSE Class 9 Maths Selina Solutions Chapter 7 PDF
ICSE Class 9 Maths Selina Solutions Chapter 7 Indices (Exponents)
Here we have provided ICSE Class 9 Maths Selina Solutions Chapter 7 – Indices (Exponent) for the ease of students so that they can prepare better for their ICSE Class 9 Maths Exam.
ICSE Class 9 Maths Selina Solutions Chapter 7 Exercise 7(A)
1. Evaluate:
(i) 3
3
x (243)
-2/3
x 9
-1/3
(ii) 5
-4
x (125)
5/3
÷ (25)
-1/2
(iii) (27/125)
2/3
x (9/25)
-3/2
(iv) 7
0
x (25)
-3/2
– 5
-3
(v) (16/81)
-3/4
x (49/9)
3/2
÷ (343/216)
2/3
Solution:
(i) 3
3
x (243)
-2/3
x 9
-1/3
= 3
3
x (3 x 3 x 3 x 3 x 3)
-2/3
x (3 x 3)
-1/3
= 3
3
x (3
5
)
-2/3
x (3
2
)
-1/3
= 3
3
x (3)
-10/3
x 3
-2/3
[As (a
m
)
n
= a
mn
]
= 3
3 – 10/3 – 2/3
[a
m
x a
n
x a
p
= a
m + n + p
]
= 3
(9 – 10 – 2)/3
= 3
-3/3
= 3
-1
= 1/3
(ii) 5
-4
x (125)
5/3
÷ (25)
-1/2
= 5
-4
x (5 x 5 x 5)
5/3
÷ (5 x 5)
-1/2
= 5
-4
x (5
3
)
5/3
÷ (5
2
)
-1/2
= 5
-4
x (5
3 x 5/3
) ÷ (5
2 x -1/2
) [As (a
m
)
n
= a
mn
]
= 5
-4
x 5
5
÷ 5
-1
= 5
-4
x 5
5
x 5
-(-1)
[As 1/a
-m
= a
-(-m)
= a
m
]
= 5
(-4 + 5 + 1)
[a
m
x a
n
x a
p
= a
m + n + p
]
= 5
2
= 25
(iii) (27/125)
2/3
x (9/25)
-3/2
= (3 x 3 x 3/5 x 5 x 5)
2/3
x (3 x 3/5 x 5)
-3/2
= (3
3
/5
3
)
2/3
x (3
2
/5
2
)
-3/2
= (3/5)
3 x 2/3
x (3/5)
2 x -3/2
[As (a
m
)
n
= a
mn
]
= (3/5)
2
x (3/5)
-3
= (3/5)
2 – 3
[a
m
x a
n
= a
m + n
]
= (3/5)
-1
= 5/3
(iv) 7
0
x (25)
-3/2
– 5
-3
= 1 x (5 x 5)
-3/2
– 5
-3
[As a
0
= 1]
= (5
2
)
-3/2
– 5
-3
= (5)
2 x -3/2
– 5
-3
[As (a
m
)
n
= a
mn
]
= 5
-3
– 5
-3
= 1/5
3
– 1/5
3
= 0
(v) (16/81)
-3/4
x (49/9)
3/2
÷ (343/216)
2/3
= (2 x 2 x 2 x 2/3 x 3 x 3 x 3)
-3/4
x (7 x 7/3 x 3)
3/2
÷ (7 x 7 x 7/6 x 6 x 6)
2/3
= (2
4
/3
4
)
-3/4
x (7
2
/3
2
)
3/2
÷ (7
3
/6
3
)
2/3
= (2/3)
4 x -3/4
x (7/3)
2 x 3/2
÷ (7/6)
3 x 2/3
= (2/3)
-3
x (7/3)
3
÷ (7/6)
2
[As (a
m
)
n
= a
mn
]
= [1/(2/3)
3
x (7/3)
3
]/ (7/6)
2
[As a
-m
= 1/a
m
]
= [(3/2)
3
x (7/3)
3
] x (7/6)
-2
= (3/2)
3
x (7/3)
3
x (6/7)
2
[As (a/b)
-m
= (b/a)
m
]
= 3/2 x 3/2 x 3/2 x 7/3 x 7/3 x 7/3 x 6/7 x 6/7
= (7 x 3 x 3)/2
= 63/2
= 31.5
2. Simplify:
(i) (8x
3
÷ 125y
3
)
2/3
(ii) (a + b)
-1
. (a
-1
+ b
-1
)
(iii)
(iv) (3x
2
)
-3
×
(x
9
)
2/3
Solution:
(i) (8x
3
÷ 125y
3
)
2/3
= (8x
3
/125y
3
)
2/3
= (2x × 2x × 2x/5y × 5y × 5y)
2/3
= (2x
3
/5y
3
)
2/3
= (2x/5y)
3 x 2/3
= (2x/5y)
2
= 4x
2
/25y
2
(ii) (a + b)
-1
. (a
-1
+ b
-1
) = 1/(a + b) x (1/a + 1/b)
= 1/(a + b) x (b + a)/ab
= 1/ab
(iii)
= (5
1
x 19)/5
= 19
(iv) (3x
2
)
-3
× (x
9
)
2/3
= 3
-3
× (x
2
)
-3
× (x)
9 x 2/3
= 3
-3
× (x)
2 x -3
× (x)
9 x 2/3
= 3
-3
× x
-6
× x
6
= 3
-3
x 1
= 1/27
3. Evaluate:
(i) √¼ + (0.01)
-1/2
– (27)
2/3
(ii) (27/8)
2/3
– (1/4)
-2
+ 5
0
Solution:
(i) √¼ + (0.01)
-1/2
– (27)
2/3
= √(½ x ½) + (0.1 x 0.1)
-1/2
– (3 x 3 x 3)
2/3
= √(½)
2
+ (0.1
2
)
-1/2
– (3
3
)
2/3
= ½ + (0.1)
2 x -1/2
– (3)
3 x
2/3
= ½ + (0.1)
-1
– (3)
2
= ½ + 1/0.1 – 3
2
= ½ + 1/(1/10) – 9
= ½ + 10 – 9
= ½ + 1
= 3/2
(ii)
(27/8)
2/3
– (1/4)
-2
+ 5
0
= (3 x 3 x 3/2 x 2 x 2)
2/3
– (1/2 x 1/2)
-2
+ 5
0
= (3
3
/2
3
)
2/3
– (½)
2
)
-2
+ 1
= (3/2)
3 x 2/3
– (½)
-4
+ 1
= (3/2)
2
– (½)
-4
+ 1
= (3/2)
2
– 2
4
+ 1
= (3 x 3)/(2 x 2) – (2 x 2 x 2 x 2) + 1
= 9/4 – 16 + 1
= (9 – 64 + 4)/4
= -51/4
4. Simplify each of the following and express with positive index:
(i) (3
-4
/2
-8
)
1/4
(ii) (27
-3
/9
-3
)
1/5
(iii) (32)
-2/5
÷ (125)
-2/3
(iv) [1 – {1 – (1 – n)
-1
}
-1
]
-1
Solution:
(i) (3
-4
/2
-8
)
1/4
= (2
8
/3
4
)
1/4
= (2
8
)
1/4
/(3
4
)
1/4
= (2)
8/4
/(3)
4/4
= 2
2
/3
= 4/3
(ii) (27
-3
/9
-3
)
1/5
= (9
3
/27
3
)
1/5
= [(3 x 3)
3
/(3 x 3 x 3)
3
]
1/5
= [(3
2
)
3
/(3
3
)
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
= 1/3
3/5
(iii) (32)
-2/5
÷ (125)
-2/3
= (32)
-2/5
/(125)
-2/3
= (125)
2/3
/(32)
2/5
= (5 x 5 x 5)
2/3
/(2 x 2 x 2 x 2 x 2)
2/5
= (5
3
)
2/3
/(2
5
)
2/5
= 5
3 x 2/3
/2
5 x 2/5
= 5
2
/2
2
= 25/4
(iv) [1 – {1 – (1 – n)
-1
}
-1
]
-1
= [1 – {1 – 1/(1 – n)}
-1
]
-1
= [1 – {((1 – n) – 1)/(1 – n)}
-1
]
-1
= [1 – {– n/(1 – n)}
-1
]
-1
= [1 – {– (1 – n)/n}]
-1
= [1 + (1 – n)/n]
-1
= [(n + 1 – n)/n]
-1
= [1/n]
-1
= n
5. If 2160 = 2
a
. 3
b
. 5
c
, find a, b and c. Hence, calculate the value of 3
a
x 2
-b
x 5
-c
.
Solution:
We have,
2160 = 2
a
x 3
b
x 5
c
(2 x 2 x 2 x 2) x (3 x 3 x 3) x 5 = 2
a
x 3
b
x 5
c
2
4
x 3
3
x 5
1
= 2
a
x 3
b
x 5
c
⇒ 2
a
x 3
b
x 5
c
= 2
4
x 3
3
x 5
1
Comparing the exponents of 2, 3 and 5 on both sides, we get
a = 4, b = 3 and c = 1
Hence, the value
3
a
x 2
-b
x 5
-c
= 3
4
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
6. If 1960 = 2
a
. 5
b
. 7
c
, calculate the value of 2
-a
. 7
b
. 5
-c
.
Solution:
We have,
1960 = 2
a
x 5
b
x 7
c
(2 x 2 x 2) x 5 x (7 x 7) = 2
a
x 5
b
x 7
c
2
3
x 5
1
x 7
2
= 2
a
x 5
b
x 7
c
⇒ 2
a
x 5
b
x 7
c
= 2
2
x 5
1
x 7
2
Comparing the exponents of 2, 5 and 7 on both sides, we get
a = 3, b = 1 and c = 2
Hence, the value
2
-a
. 7
b
. 5
-c
= 2
-3
x 7
1
x 5
-2
= (½ x ½ x ½) x 7 x (1/5 x 1/5)
= 1/8 x 7 x 1/25
= 7/200
7. Simplify:
(i)
(ii)
Solution:
(i)
(ii)
8. Show that:
(a
m
/a
-n
)
m-n
× (a
n
/a
-l
)
n-l
× (a
l
/a
-m
)
l-m
= 1
Solution:
Taking the L.H.S., we have
(a
m
/a
-n
)
m-n
× (a
n
/a
-l
)
n-l
× (a
l
/a
-m
)
l-m
= (a
m
× a
n
)
m-n
× (a
n
×a
l
)
n-l
× (a
l
× a
m
)
l-m
= (a
m+n
)
m-n
× (a
n+l
)
n-l
× (a
l+m
)
l-m
= a
0
= 1
9. If a = x
m+n
. x
l
; b = x
n+l
.
x
m
and c = x
l+m
. x
n
,
Prove that: a
m-n
. b
n-l
. c
l-m
= 1
Solution:
We have,
a = x
m+n
. x
l
b = x
n+l
.
x
m
c = x
l+m
. x
n
Now,
Considering the L.H.S.,
a
m-n
. b
n-l
. c
l-m
= (x
m+n
. x
l
)
m-n
. (x
n+l
.
x
m
)
n-l
. (x
l+m
. x
n
)
l-m
= [x
(m+n)(m-n)
. x
l(m-n)
]. [x
(n+l)(n-l)
. x
m(n-l)
]. [x
(l+m)(l-m)
. x
n(l-m)
]
= x
0
= 1
= R.H.S
– Hence proved.
10. Simplify:
(i)
(ii)
Solution:
(i)
= x
0
= 1
(ii)
Exercise 7(B)
1. Solve for x:
(i) 2
2x+1
= 8
(ii) 2
5x-1
= 4 ×
2
3x + 1
(iii) 3
4x + 1
= (27)
x + 1
(iv) (49)
x + 4
= 7
2
× (343)
x + 1
Solution:
(i) We have, 2
2x+1
= 8
⇒2
2x+1
= 2
3
Now, if the bases are equal, then the powers must be equal
On comparing the exponents, we get
2x + 1 = 3
2x = 3 – 1
2x = 2
x = 2/2
x = 1
(ii) We have, 2
5x-1
= 4 ×
2
3x + 1
⇒ 2
5x-1
= 2
2
×
2
3x + 1
2
5x-1
= 2
(3x + 1) + 2
2
5x-1
=
2
3x + 3
Now, if the bases are equal, then the powers must be equal
On comparing the exponents, we get
5x – 1 = 3x + 3
5x – 3x = 3 + 1
2x = 4
x = 4/2
x = 2
(iii) We have, 3
4x + 1
= (27)
x + 1
3
4x + 1
= (3
3
)
x + 1
3
4x + 1
= (3)
3x + 3
Now, if the bases are equal, then the powers must be equal
On comparing the exponents, we get
4x + 1 = 3x + 3
4x – 3x = 3 – 1
x = 2
(iv) We have, (49)
x + 4
= 7
2
× (343)
x + 1
(7 x 7)
x + 4
= 7
2
× (7 x 7 x 7)
x + 1
(7
2
)
x + 4
= 7
2
× (7
3
)
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
2. Find x, if:
(i) 4
2x
= 1/32
(ii) √(2
x+3
) = 16
(iii) [√(⅗)]
x+1
= 125/27
(iv) [∛(⅔)]
x-1
= 27/8
Solution:
(i) We have, 4
2x
= 1/32
(2 × 2)
2x
= 1/(2 × 2 × 2 × 2 × 2)
(2
2
)
2x
= 1/2
5
2
4x
= 2
-5
Now, if the bases are equal, then the powers must be equal
So, on comparing the exponents, we get
4x = -5
x = -5/4
(ii) We have, √(2
x+3
) = 16
(2
x+3
)
1/2
= (2 × 2 × 2 × 2)
2
(x+3)/2
= 2
4
Now, if the bases are equal, then the powers must be equal
So, on comparing the exponents, we get
(x + 3)/2 = 4
x + 3 = 8
x = 8 – 3
x = 5
(iii) We have, [√(⅗)]
x+1
= 125/27
[(⅗)
1/2
]
x+1
= (5 × 5 × 5)/(3 × 3 × 3)
(⅗)
(x+1)/2
= 5
3
/3
3
(⅗)
(x+1)/2
= (5/3)
3
(⅗)
(x+1)/2
= (3/5)
-3
Now, if the bases are equal, then the powers must be equal
So, on comparing the exponents, we get
(x + 1)/2 = -3
x + 1 = -6
x = -6 – 1
x = -7
(iv) We have, [∛(⅔)]
x-1
= 27/8
[(⅔)
1/3
]
x-1
= (3 × 3 × 3)/(2 × 2 × 2)
(⅔)
(x-1)/3
= (3/2)
3
(⅔)
(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
3. Solve:
(i) 4
x-2
– 2
x+1
= 0
(ii)
Solution:
(i) We have,
4
x-2
– 2
x+1
= 0
(2
2
)
x-2
– 2
x+1
= 0
2
2x-4
– 2
x+1
= 0
2
2x-4
= 2
x+1
Now, if the bases are equal, then the powers must be equal
So, on comparing the exponents, we get
2x – 4 = x + 1
2x – x = 4 + 1
x = 5
(ii) We have,
Now, if the bases are equal, then the powers must be equal
So, on comparing the exponents, we get
x
2
= x+ 2
x
2
– x – 2 = 0
On factorization, we get
x
2
– 2x + x – 2 = 0
x(x – 2 + 1(x – 2) = 0
(x + 1)(x – 2) = 0
So, either (x + 1) = 0 or (x – 2) = 0
Thus, x = -1 or 2
4. Solve:
(i) 8 × 2
2x
+ 4
× 2
x+1
= 1 + 2
x
(ii) 2
2x
+ 2
x+2
– 4 × 2
3
= 0
(iii) (√3)
x-3
= (∜3)
x+1
Solution:
(i) We have, 8 × 2
2x
+ 4
× 2
x+1
= 1 + 2
x
8 × (2
x
)
2
+ 4
× (2
x
) × 2
1
= 1 + 2
x
Let us substitute 2
x
= t
Then,
8 × t
2
+ 4
× t × 2 = 1 + t
8t
2
+ 8t = 1 + t
8t
2
+ 8t – t – 1 = 0
8t
2
+ 7t – 1 = 0
8t
2
+ 8t – t – 1 = 0
8t(t + 1) – 1(t + 1) = 0
(8t – 1)(t + 1) = 0
So, either 8t – 1 = 0 or t + 1 = 0
Thus, t = 1/8 or -1
Now, we have
2
x
= t
So,
2
x
= 1/8 or 2
x
= -1
The equation, 2
x
= -1 is not possible
Hence, for 2
x
= 1/8
2
x
= 1/(2 × 2 × 2)
2
x
= 1/2
3
2
x
= 2
-3
Now, if the bases are equal, then the powers must be equal
So, on comparing the exponents, we get
x = -3
(ii) We have,
2
2x
+ 2
x+2
– 4 × 2
3
= 0
2
2x
+ 2
x+2
– 2
2
× 2
3
= 0
(2
x
)
2
+ 2
x
.2
2
– 2
3+2
= 0
(2
x
)
2
+ 2
x
.2
2
– 2
5
= 0
Now, let’s assume 2
x
= t
So, the above equation becomes
(t)
2
+ t.2
2
– 2
5
= 0
t
2
+ 4t – 32 = 0
On factorization, we have
t
2
+ 8t – 4t – 32 = 0
t(t + 8) – 4(t + 8) = 0
(t – 4)(t + 8) = 0
So, either (t – 4) = 0 or (t + 8) = 0
Thus, t = 4 or -8
Now, we have t = 2
x
So,
2
x
= 4 or 2
x
= -8
The equation, 2
x
= -8 is not possible
Hence, for
2
x
= 4
2
x
= 2
2
On comparing the exponents, we get
x = 2
(iii) (√3)
x-3
= (∜3)
x+1
(3
1/2
)
x-3
= (3
1/4
)
x+1
3
(x-3)/2
= 3
(x+1)/4
Now, if the bases are equal, then the powers must be equal
So, on comparing the exponents, we get
(x – 3)/2 = (x + 1)/4
2(x – 3) = (x + 1)
2x – 6 = x + 1
2x – x = 6 + 1
x = 7
5. Find the values of m and n if:
4
2m
= (∛16)
-6/n
= (√8)
2
Solution:
We have, 4
2m
= (∛16)
-6/n
= (√8)
2
Now, considering
4
2m
= (√8)
2
(2
2
)
2m
= (8
1/2
)
2
2
4m
= 8
1/2 x 2
2
4m
= 8
⇒ 2
4m
= 2
3
Now, if the bases are equal, then the powers must be equal
So, on comparing the exponents, we get
4m = 3
m = ¾
Now, from the given considering
(∛16)
-6/n
= (√8)
2
(16
1/3
)
-6/n
= (8
1/2
)
2
(16)
1/3 x -6/n
= 8
1/2 x 2
(16)
-2/n
= 8
(2 x 2 x 2 x 2)
-2/n
= (2 x 2 x 2)
(2)
4 x -2/n
= 2
3
(2)
-8/n
= 2
3
Now, if the bases are equal, then the powers must be equal
So, on comparing the exponents, we get
-8/n = 3
n = -8/3
Therefore, the value of m and n are ¾ and -8/3
6. Solve x and y if:
(√32)
x
÷ 2
y+1
= 1 and 8
y
– 16
4 – x/2
= 0
Solution:
Consider the equation, (√32)
x
÷ 2
y+1
= 1
(√(2 × 2 × 2 × 2 × 2))
x
÷ 2
y+1
= 1
(√2
5
)
x
÷ 2
y+1
= 1
(2
5
)
1/2 × x
÷ 2
y+1
= 1
2
5x/2
÷ 2
y+1
= 1
(2
5x/2
)/(2
y+1
) = 1
2
5x/2 – (y+1)
= 2
0
Now, if the bases are equal, then the powers must be equal
So, on comparing the exponents, we get
5x/2 – (y+1) = 0
5x/2 – y – 1 = 0
5x – 2y – 2 = 0 … (i)
Next, let’s consider 8
y
– 16
4 – x/2
= 0
(2 × 2 × 2)
y
– (2 × 2 × 2 × 2)
4 – x/2
= 0
(2
3
)
y
– (2
4
)
4 – x/2
= 0
2
3y
– 2
16 – 2x
= 0
2
3y
= 2
16 – 2x
Now, if the bases are equal, then the powers must be equal
So, on comparing the exponents, we get
3y = 16 – 2x
2x + 3y – 16 = 0 … (ii)
On solving equations (i) and (ii),
By manipulating by (i) × 3 + (ii) × 2, we have
15x – 6y – 6 = 0
4x + 6y – 32 = 0
———————–
19x – 38 = 0
x = 38/19
x = 2
Now, substituting the value of x in (i)
5(2) – 2y – 2 = 0
10 – 2y – 2 = 0
8 = 2y
y = 8/2
y = 4
Therefore, the values of x and y are 2 and 4 respectively
7. Prove that:
(i) (x
a
/x
b
)
a+b-c
. (x
b
/x
c
)
b+c-a
. (x
c
/x
a
)
c+a-b
= 1
(ii) x
a(b-c)
/x
b(a-c)
÷ (x
b
/x
a
)
c
= 1
Solution:
(i) Taking L.H.S, we have
(x
a
/x
b
)
a+b-c
. (x
b
/x
c
)
b+c-a
. (x
c
/x
a
)
c+a-b
= (x
a
/x
b
)
a+b-c
. (x
b
/x
c
)
b+c-a
. (x
c
/x
a
)
c+a-b
= x
(a-b)(a+b-c)
. x
(b-c)(b+c-a)
. x
(c-a)(c+a-b)
= x
0
= 1
= R.H.S
(ii) Taking L.H.S, we have
x
a(b-c)
/x
b(a-c)
÷ (x
b
/x
a
)
c
= x
a(b-c) – b(a-c)
÷ x
c(b-a)
= x
a(b-c) – b(a-c)
/x
c(b-a)
= x
a(b-c) – b(a-c) – c(b-a)
= x
ab-ac-ba+bc-cb+ac
= x
0
= 1
= R.H.S
8. If a
x
= b, b
y
= c and c
z
= a, prove that: xyz = 1.
Solution:
We have, a
x
= b, b
y
= c and c
z
= a
Now, considering
a
x
= b
On raising to the power yz on both sides, we get
(a
x
)
yz
= (b)
yz
(a)
xyz
= (b
y
)
z
(a)
xyz
= (c)
z
[As, b
y
= c]
a
xyz
= a
a
xyz
= a
1
[As, c
z
= a]
Now, if the bases are equal, then the powers must be equal
On comparing the exponents, we get
Hence, xyz = 1
9. If a
x
= b
y
= c
z
and b
2
= ac, prove that: y = 2az/(x + z).
Solution:
Let’s assume a
x
= b
y
= c
z
= k
So,
a = k
1/x
; b = k
1/y
and c = k
1/z
Now,
It’s given that b
2
= ac
⇒ (k
1/y
)
2
= (k
1/x
) × (k
1/z
)
(k
2/y
) = k
1/x + 1/z
k
2/y
= k
(z+x)/xz
Now, if the bases are equal, then the powers must be equal
On comparing the exponents, we get
2/y = (z + x)/xz
2xz = y(z + x)
Hence,
y = 2xz/(x + z)
10. If 5
-p
= 4
-q
= 20
r
, show that: 1/p + 1/q + 1/r = 0.
Solution:
Let’s assume 5
-p
= 4
-q
= 20
r
= k
Then, as
5
-p
= k ⇒ 5 = k
-1/p
4
-q
= k ⇒ 4 = k
-1/q
20
r
= k ⇒ 20 = k
1/r
Now, we know
5 x 4 = 20
(k
-1/p
)
x (k
-1/q
) = k
1/r
k
-1/p – 1/q
= k
1/r
Now, if the bases are equal, then the powers must be equal
On comparing the exponents, we get
-1/p – 1/q = 1/r
Hence,
1/p + 1/q + 1/r = 0
11. If m ≠ n and (m + n)
-1
(m
-1
+ n
-1
) = m
x
n
y
, show that: x + y + 2 = 0
Solution:
Given equation,
(m + n)
-1
(m
-1
+ n
-1
) = m
x
n
y
1/(m + n) × (1/m + 1/n) = m
x
n
y
1/(m + n) × (m + n)/mn = m
x
n
y
1/mn = m
x
n
y
m
-1
n
-1
= m
x
n
y
Now, if the bases are equal, then the powers must be equal
On comparing the exponents, we get
x = -1 and y = -1
Substituting the values of x and y in the equation x + y + 2 = 0, we have
(-1) + (-1) + 2 = 0
0 = 0
L.H.S = R.H.S
Therefore, x + y + 2 = 0
12. If 5
x + 1
= 25
x – 2
, find the value of 3
x – 3
× 2
3 – x
Solution:
We have, 5
x + 1
= 25
x – 2
5
x + 1
= (5
2
)
x – 2
5
x + 1
= 5
2x – 4
Now, if the bases are equal, then the powers must be equal
On comparing the exponents, we get
x + 1 = 2x – 4
2x – x = 4 + 1
x = 5
Hence, the value of
3
x – 3
× 2
3 – x
= 3
5 – 3
× 2
3 – 5
= 3
2
× 2
-2
= 9 × ¼
= 9/4
13. If 4
x + 3
= 112 + 8 × 4
x
, find the value of (18x)
3x
.
Solution:
We have,
4
x + 3
= 112 + 8 × 4
x
4
x
.4
3
= 112 + 8 × 4
x
Let’s assume 4
x
= t
Then,
t
.4
3
= 112 + 8 × t
64t = 112 + 8t
64t – 8t = 112
56t = 112
t = 112/56
t = 2
But we have taken 4
x
= t
So, 4
x
= 2
(2
2
)
x
= 2
1
2
2x
= 2
1
Now, if the bases are equal, then the powers must be equal
On comparing the exponents, we get
2x = 1
x = ½
Now, the value of (18x)
3x
will be
= (18 × ½)
3 × 1/2
= (9)
3/2
= (3
2
)
3/2
= 3
3
= 27
14. Solve for x:
(i) 4
x-1
× (0.5)
3 – 2x
= (1/8)
-x
Solution:
We have,
4
x-1
× (0.5)
3 – 2x
= (1/8)
-x
(2
2
)
x-1
× (1/2)
3 – 2x
= (1/2
3
)
-x
(2)
2x-1
× (2)
-(3 – 2x)
= (2
-3
)
-x
(2)
2x-2
× (2)
2x-3
= (2)
3x
2
(2x-2) + (2x-3)
= (2)
3x
2
4x-5
= 2
3x
Now, if the bases are equal, then the powers must be equal
On comparing the exponents, we get
4x – 5 = 3x
4x – 3x = 5
x = 5
(ii) (a
3x + 5
)
2
×
(a
x
)
4
= a
8x + 12
Solution:
We have,
(a
3x + 5
)
2
× (a
x
)
4
= a
8x + 12
a
6x + 10
× a
4x
= a
8x + 12
a
6x + 10 + 4x
= a
8x + 12
a
10x + 10
= a
8x + 12
Now, if the bases are equal, then the powers must be equal
On comparing the exponents, we get
10x + 10 = 8x + 12
10x – 8x = 12 – 10
2x = 2
x = 1
(iii) (81)
3/4
– (1/32)
-2/5
+ x(1/2)
-1
× 2
0
= 27
Solution:
We have,
(81)
3/4
– (1/32)
-2/5
+ x(1/2)
-1
× 2
0
= 27
(3
4
)
3/4
– (1/2
5
)
-2/5
+ x(1/2)
-1
× 2
0
= 27
(3
4
)
3/4
– (2
-5
)
-2/5
+ x(2
-1
)
-1
× 2
0
= 27
3
3
– 2
2
+ 2x × 1 = 27
27 – 4 + 2x = 27
2x + 23 = 27
2x = 27 – 23
2x = 4
x = 4/2
x = 2
(iv) 2
3x + 3
= 2
3x + 1
+ 48
Solution:
We have,
2
3x + 3
= 2
3x + 1
+ 48
2
3x + 3
– 2
3x + 1
= 48
2
3x
(2
3
– 2
1
) = 48
2
3x
(8 – 2) = 48
2
3x
× 6 = 48
2
3x
= 48/6
2
3x
= 8
2
3x
= 2
3
Now, if the bases are equal, then the powers must be equal
On comparing the exponents, we get
3x = 3
x = 1
(v) 3(2
x
+ 1) – 2
x+2
+ 5 = 0
Solution:
We have,
3(2
x
+ 1) – 2
x+2
+ 5 = 0
3 × 2
x
+ 3 – 2
x
.2
2
+ 5 = 0
2
x
(3 – 2
2
) + 5 + 3 = 0
2
x
(3 – 4) + 8 = 0
-2
x
+ 8 = 0
2
x
= 8
2
x
= 2
3
Now, if the bases are equal, then the powers must be equal
On comparing the exponents, we get
x = 3
(vi) 9
x+2
= 720 + 9
x
Solution:
We have,
9
x+2
= 720 + 9
x
9
x+2
– 9
x
= 720
9
x
(9
2
– 1) = 720
9
x
(81 – 1) = 720
9
x
(80) = 720
9
x
= 9
9
x
= 9
1
Therefore, x = 1
Exercise 7(C)
1. Evaluate:
(i) 9
5/2
– 3 x 8
0
– (1/81)
-1/2
(ii) (64)
2/3
– ∛125 – 1/2
-5
+ (27)
-2/3
x (25/9)
-1/2
(iii) [(-2/3)
-2
]
3
x (1/3)
-4
x 3
-1
x 1/6
Solution:
(i) We have, 9
5/2
– 3 x 8
0
– (1/81)
-1/2
= (3
2
)
5/2
– 3 x 1 – (1/9
2
)
-1/2
= 3
2 x 5/2
– 3 – (1/9)
-2 x ½
= 3
5
– 3 – (1/9)
-1
= (3 x 3 x 3 x 3 x 3) – 3 – (9
-1
)
-1
= 243 – 3 – 9
= 231
(ii) We have, (64)
2/3
– ∛125 – 1/2
-5
+ (27)
-2/3
x (25/9)
-1/2
= (4 x 4 x 4)
2/3
– (5 x 5 x 5)
1/3
– 1/2
-5
+ (3 x 3 x 3)
-2/3
x (5 x 5/3 x 3)
-1/2
= (4
3
)
2/3
– (5
3
)
1/3
– (2
-1
)
-5
+ (3
3
)
-2/3
x (5
2
/3
2
)
-1/2
= (4)
3 x 2/3
– (5)
3 x 1/3
– (2)
-1 x -5
+ (3)
3 x -2/3
x (5/3)
2 x -1/2
= (4)
2
– 5 – 2
5
+ 3
-2
x (5/3)
-1
= 16 – 5 – 32 + 1/9 x 3/5
= -21 + 1/15
= (-315 + 1)/15
= -314/15
(iii) We have, [(-2/3)
-2
]
3
x (1/3)
-4
x 3
-1
x 1/6
= (-2/3)
-6
x (3
-1
)
-4
x 3
-1
x ½ x 1/3
= (-3/2)
6
x 3
4
x 3
-1
x 2
-1
x 3
-1
= (-1)
6
x (3
6
x 3
4
x 3
-1
x 3
-1
) x (2
-6
x 2
-1
)
= 1 x 3
6+4-1-1
x 2
-6-1
= 3
8
x 2
-7
= 3
8
/2
7
2. Simplify:
Solution:
We have,
= (27 – 9)/(81 – 9)
= 18/72
= 1/4
3. Solve: 3
x-1
× 5
2y-3
= 225.
Solution:
Given, 3
x-1
× 5
2y-3
= 225
3
x-1
× 5
2y-3
= 9 x 25
3
x-1
× 5
2y-3
= 3
2
x 5
2
Now, if the bases are equal, then the powers must be equal
On comparing the exponents, we get
x – 1 = 2 and 2y – 3 = 2
x = 2 + 1 and 2y = 2 + 3
x = 3 and 2y = 5
x = 3 and y = 5/2 = 2.5
4. If [(a
-1
b
2
)/(a
2
b
-4
)]
7
÷ [(a
3
b
-5
)/(a
-2
b
3
)] = a
x
.b
y
, find x + y.
Solution:
We have,
(b
42
/a
21
)
÷ (b
40
/a
25
) = a
x
. b
y
(b
42
/a
21
)
× (a
25
/b
40
) = a
x
. b
y
b
42 – 40
× a
25 – 21
= a
x
. b
y
a
4
× b
2
= a
x
. b
y
Now, if the bases are equal, then the powers must be equal
On comparing the exponents, we get
x = 4 and y = 2
x + y = 4 + 2 = 6
5. If 3
x+1
= 9
x-3
, find the value of 2
1+x
.
Solution:
We have,
3
x+1
= 9
x-3
3
x
× 3
1
= (3
2
)
x – 3
3
x
× 3
1
= (3)
2x – 6
3
x
= (3)
2x – 6
/3
3
x
= (3)
2x – 6
× 3
-1
3
x
= (3)
2x – 6 – 1
3
x
= 3
2x – 7
Now, if the bases are equal, then the powers must be equal
On comparing the exponents, we get
x = 2x – 7
x = 7
Now,
2
1+x
= 2
1+7
= 2
8
= 256
6. If 2
x
= 4
y
= 8
z
and 1/2x + 1/4y + 1/8z = 4, find the value of x.
Solution:
Given, 2
x
= 4
y
= 8
z
2
x
= (2
2
)
y
= (2
3
)
z
2
x
= 2
2y
= 2
3z
On comparing the powers, we get
x = 2y = 3z
y = x/2 and z = x/3
Now,
1/2x + 1/4y + 1/8z = 4
1/2x + 1/4(x/2) + 1/8(x/3) = 4
1/2x + 2/4x + 3/8x = 4
1/2x + 1/2x + 3/8x = 4
(4 + 4 + 3)/8x = 4
11/8x = 4
4 × 8x = 11
32x = 11
x = 11/32
7. If
Show that: m – n = 1
Solution:
We have,
On comparing the powers, we get
m – n = 1
8. Solve for x: (13)
√x
= 4
4
– 3
4
– 6.
Solution:
We have,
(13)
√x
= 4
4
– 3
4
– 6
= 256 – 81 – 6
= 169
= 13
2
13
√x
= 13
2
On comparing the powers, we get
√x = 2
Squaring on both sides,
x = 2
2
x = 4
9. If 3
4x
= (81)
-1
and (10)
1/y
= 0.0001, find the value of 2
-x
×
16
y
.
Solution:
We have, 3
4x
= (81)
-1
and (10)
1/y
= 0.0001
3
4x
= (3
4
)
-1
and (10)
1/y
= 1/10000
3
4x
= 3
-4
and 10
1/y
= 1/10
-4
3
4x
= 3
-4
and 10
1/y
= 1/10
4
3
4x
= 3
-4
and 10
1/y
= 10
-4
On comparing the powers, we get
4x = -4 and 1/y = -4
x = -1 and y = -1/4
Now, value of
2
-x
×
16
y
= 2
-(-1)
×
16
-1/4
= 2
1
×
(2
4
)
-1/4
= 2 x 2
-1
= 2/2
= 1
10. Solve: 3(2
x
+ 1) – 2
x+2
+ 5 = 0.
Solution:
We have, 3(2
x
+ 1) – 2
x+2
+ 5 = 0
(3 × 2
x
+ 3) – (2
x
× 2
2
) + 5 = 0
2
x
(3 – 2
2
) + 3 + 5 = 0
2
x
(3 – 4) + 8 = 0
2
x
(-1) + 8 = 0
-2
x
+ 8 = 0
2
x
= 8
2
x
= 2
3
On comparing the powers, we get
x = 3
11. If (a
m
)
n
= a
m
.a
n
, find the value of: m(n – 1) – (n – 1)
Solution:
We have, (a
m
)
n
= a
m
.a
n
a
mn
= a
m+n
On comparing the powers, we get
mn = m + n … (i)
Now,
m(n – 1) – (n – 1) = mn – m – n + 1
= (m + n) – m – n + 1 … [Form (i)]
= 1
12. If m = ∛15 and n = ∛14, find the value of m – n – 1/(m
2
+ mn + n
2
)
Solution:
We have,
m = ∛15 and n = ∛14
⇒ m = 15
3
and n = 14
3
13. Evaluate:
Solution:
We have,
= 6/225 x 25/1
= 6/9
= 2/3
14. Evaluate:
(x
q
/x
r
)
1/qr
× (x
r
/x
p
)
1/rp
× (x
p
/x
q
)
1/pq
Solution:
We have,
15. (i) Prove that: a
-1
/(a
-1
+ b
-1
) + a
-1
/(a
-1
– b
-1
) = 2b
2
/(b
2
– a
2
)
Solution:
We have,
= 1/a x ab/(b + a) + 1/a x ab/(b – a)
= b/(b + a) + b/(b – a)
= (b
2
– ab + b
2
+ ab)/(b
2
– a
2
)
= 2b
2
/(b
2
– a
2
)
= R.H.S
(ii) Prove that: (a + b + c)/(a
-1
b
-1
+ b
-1
c
-1
+ c
-1
a
-1
) = abc
Solution:
Taking L.H.S., we have
= abc
= R.H.S
16. Evaluate: 4/(216)
-2/3
+ 1/(256)
-3/4
+ 2/(243)
-1/5
Solution:
We have,
= 4/6
-2
+ 1/4
-3
+ 2/3
-1
= 4 x 6
2
+ 4
3
+ 2 x 3
1
= 4 x 36 + 64 + 2 x 3
= 144 + 64 + 6
= 214
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