Exponents and Powers - Class 7 - Mathematics
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Examples - Exponents and Powers | NCERT | Mathematics | Class 7
Express 256 as a power 2.
To express $256$ as a power of $2$, you can use the fact that $2$ raised to some power $n$ equals $256$. This means we're looking for $n$ in the equation $2^n = 256$.
The solution is $n = 8$, since $2^8 = 256$.
Therefore, $256$ expressed as a power of $2$ is $2^8$.
Which one is greater $2^{3}$ or $3^{2}$ ?
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Sign up nowWhich one is greater $8^{2}$ or $2^{8}$ ?
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Sign up nowExpand $a^{3} b^{2}, a^{2} b^{3}, b^{2} a^{3}, b^{3} a^{2}$. Are they all same?
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Sign up nowExpress the following numbers as a product of powers of prime factors:
(i) 72
(ii) 432
(iii) 1000
(iv) 16000
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Sign up nowWork out $(1)^{5},(-1)^{3},(-1)^{4},(-10)^{3},(-5)^{4}$.
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Sign up nowCan you tell which one is greater $\left(5^{2}\right) \times 3$ or $\left(5^{2}\right)^{3}$ ?
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Sign up nowExpress the following terms in the exponential form:
(i) $(2 \times 3)^{5}$
(ii) $(2 a)^{4}$
(iii) $(-4 m)^{3}$
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Sign up nowExpand:
(i) $\left(\frac{3}{5}\right)^{4}$
(ii) $\left(\frac{-4}{7}\right)^{5}$
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Sign up nowWrite exponential form for $8 \times 8 \times 8 \times 8$ taking base as 2 .
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Sign up nowSimplify and write the answer in the exponential form.
(i) $\left(\frac{3^{7}}{3^{2}}\right) \times 3^{5}$
(ii) $2^{3} \times 2^{2} \times 5^{5}$
(iii) $\left(6^{2} \times 6^{4}\right) \div 6^{3}$
(iv) $\left[\left(2^{2}\right)^{3} \times 3^{6}\right] \times 5^{6}$
(v) $8^{2} \div 2^{3}$
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Sign up nowExpress the following numbers in the standard form:
(i) 5985.3
(ii) 65,950
(iii) $3,430,000$
(iv) $70,040,000,000$
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Sign up nowExercise 11.1 - Exponents and Powers | NCERT | Mathematics | Class 7
Find the value of: with steps
(i) $2^{6}$
(ii) $9^{3}$
(iii) $11^{2}$
(iv) $5^{4}$
To find the value of each exponential expression, we use the definition of exponentiation, which is repeated multiplication of the base number by itself for the number of times indicated by the exponent.
(i) For $2^6$, the base is 2 and the exponent is 6. This means we multiply 2 by itself 6 times: $$2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$$
(ii) For $9^3$, the base is 9 and the exponent is 3. This means we multiply 9 by itself 3 times: $$9 \times 9 \times 9 = 729$$
(iii) For $11^2$, the base is 11 and the exponent is 2. This means we multiply 11 by itself 2 times: $$11 \times 11 = 121$$
(iv) For $5^4$, the base is 5 and the exponent is 4. This means we multiply 5 by itself 4 times: $$5 \times 5 \times 5 \times 5 = 625$$
Thus, the values of the given expressions are:
- $2^6 = 64$
- $9^3 = 729$
- $11^2 = 121$
- $5^4 = 625$
Express the following in exponential form:
(i) $6 \times 6 \times 6 \times 6$
(ii) $t \times t$
(iii) $b \times b \times b \times b$
(iv) $5 \times 5 \times 7 \times 7 \times 7$
(v) $2 \times 2 \times a \times a$
(vi) $a \times a \times a \times c \times c \times c \times c \times d$
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Sign up nowExpress each of the following numbers using exponential notation: with steps
(i) 512
(ii) 343
(iii) 729
(iv) 3125
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Sign up nowIdentify the greater number, wherever possible, in each of the following?
(i) $4^{3}$ or $3^{4}$
(ii) $5^{3}$ or $3^{5}$
(iii) $2^{8}$ or $8^{2}$
(iv) $100^{2}$ or $2^{100}$
(v) $2^{10}$ or $10^{2}$
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Sign up nowExpress each of the following as product of powers of their prime factors: with steps
(i) 648
(ii) 405
(iii) 540
(iv) 3,600
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Sign up nowSimplify: with steps
(i) $2 \times 10^{3}$
(ii) $7^{2} \times 2^{2}$
(iii) $2^{3} \times 5$
(iv) $3 \times 4^{4}$
(v) $0 \times 10^{2}$
(vi) $5^{2} \times 3^{3}$
(vii) $2^{4} \times 3^{2}$
(viii) $3^{2} \times 10^{4}$
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Sign up nowSimplify: with steps
(i) $(-4)^{3}$
(ii) $(-3) \times(-2)^{3}$
(iii) $(-3)^{2} \times(-5)^{2}$
(iv) $(-2)^{3} \times(-10)^{3}$
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Sign up nowCompare the following numbers: with steps
(i) $2.7 \times 10^{12} ; 1.5 \times 10^{8}$
(ii) $4 \times 10^{14} ; 3 \times 10^{17}$
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Sign up nowExercise 11.2 - Exponents and Powers | NCERT | Mathematics | Class 7
Using laws of exponents, simplify and write the answer in exponential form:
(i) $3^{2} \times 3^{4} \times 3^{8}$
(ii) $6^{15} \div 6^{10}$
(iii) $a^{3} \times a^{2}$
(iv) $7^{x} \times 7^{2}$
(v) $\left(5^{2}\right)^{3} \div 5^{3}$
(vi) $2^{5} \times 5^{5}$
(vii) $a^{4} \times b^{4}$
(viii) $\left(3^{4}\right)^{3}$
(ix) $\left(2^{20} \div 2^{15}\right) \times 2^{3}$
(x) $8^{t} \div 8^{2}$
To simplify each expression using the laws of exponents, we'll apply the following rules:
- Multiplication with same base: $a^m \times a^n = a^{m+n}$
- Division with same base: $a^m \div a^n = a^{m-n}$
- Power of a power: $(a^m)^n = a^{m \times n}$
- Multiplication of powers with different bases and same exponent: $a^m \times b^m = (a \times b)^m$
Let's simplify each expression using these rules:
(i) $$3^{2} \times 3^{4} \times 3^{8} = 3^{2+4+8} = 3^{14}$$
(ii) $$6^{15} \div 6^{10} = 6^{15-10} = 6^{5}$$
(iii) $$a^{3} \times a^{2} = a^{3+2} = a^{5}$$
(iv) $$7^{x} \times 7^{2} = 7^{x+2}$$
(v) $$\left(5^{2}\right)^{3} \div 5^{3} = 5^{2 \times 3} \div 5^{3} = 5^{6-3} = 5^{3}$$
(vi) $$2^{5} \times 5^{5} = (2 \times 5)^{5} = 10^{5}$$
(vii) $$a^{4} \times b^{4} = (a \times b)^{4}$$
(viii) $$\left(3^{4}\right)^{3} = 3^{4 \times 3} = 3^{12}$$
(ix) $$\left(2^{20} \div 2^{15}\right) \times 2^{3} = 2^{20-15+3} = 2^{8}$$
(x) $$8^{t} \div 8^{2} = 8^{t-2}$$
These simplifications demonstrate the power of exponent laws in combining and simplifying expressions.
Simplify and express each of the following in exponential form:
(i) $\frac{2^{3} \times 3^{4} \times 4}{3 \times 32}$
(ii) $\left(\left(5^{2}\right)^{3} \times 5^{4}\right) \div 5^{7}$
(iii) $25^{4} \div 5^{3}$
(iv) $\frac{3 \times 7^{2} \times 11^{8}}{21 \times 11^{3}}$
(v) $\frac{3^{7}}{3^{4} \times 3^{3}}$
(vi) $2^{0}+3^{0}+4^{0}$
(vii) $2^{0} \times 3^{0} \times 4^{0}$
(viii) $\left(3^{0}+2^{0}\right) \times 5^{0}$
(ix) $\frac{2^{8} \times a^{5}}{4^{3} \times a^{3}}$
(x) $\left(\frac{a^{5}}{a^{3}}\right) \times a^{8}$
(xi) $\frac{4^{5} \times a^{8} b^{3}}{4^{5} \times a^{5} b^{2}}$
(xii) $\left(2^{3} \times 2\right)^{2}$
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Sign up nowSay true or false and justify your answer:
(i) $10 \times 10^{11}=100^{11}$
(ii) $2^{3}>5^{2}$
(iii) $2^{3} \times 3^{2}=6^{5}$
(iv) $3^{0}=(1000)^{0}$
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Sign up nowExpress each of the following as a product of prime factors only in exponential form:
(i) $108 \times 192$
(ii) 270
(iii) $729 \times 64$
(iv) 768
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Sign up nowSimplify with steps
(i) $\frac{\left(2^{5}\right)^{2} \times 7^{3}}{8^{3} \times 7}$
(ii) $\frac{25 \times 5^{2} \times t^{8}}{10^{3} \times t^{4}}$
(iii) $\frac{3^{5} \times 10^{5} \times 25}{5^{7} \times 6^{5}}$
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Sign up nowExercise 11.3 - Exponents and Powers | NCERT | Mathematics | Class 7
Write the following numbers in the expanded forms:
279404, 3006194, 2806196, 120719, 20068
To write a number in its expanded form, we start from the left-most digit and move towards the right, expressing each digit as its value in that place. For example, in the number 279404, the digit 2 is in the hundred thousands place, so its value is (2 \times 100,000), the digit 7 is in the ten thousands place, so its value is (7 \times 10,000), and so on.
I'll perform this expansion for each of the provided numbers.
-
279404:
- (2 \times 100,000 + 7 \times 10,000 + 9 \times 1,000 + 4 \times 100 + 0 \times 10 + 4)
-
3006194:
- (3 \times 1,000,000 + 0 \times 100,000 + 0 \times 10,000 + 6 \times 1,000 + 1 \times 100 + 9 \times 10 + 4)
-
2806196:
- (2 \times 1,000,000 + 8 \times 100,000 + 0 \times 10,000 + 6 \times 1,000 + 1 \times 100 + 9 \times 10 + 6)
-
120719:
- (1 \times 100,000 + 2 \times 10,000 + 0 \times 1,000 + 7 \times 100 + 1 \times 10 + 9)
-
20068:
- (2 \times 10,000 + 0 \times 1,000 + 0 \times 100 + 6 \times 10 + 8)
These are the numbers written in their expanded forms, showing how each digit contributes to the overall number based on its place value.
Find the number from each of the following expanded forms:
(a) $8 \times 10^{4}+6 \times 10^{3}+0 \times 10^{2}+4 \times 10^{1}+5 \times 10^{0}$
(b) $4 \times 10^{5}+5 \times 10^{3}+3 \times 10^{2}+2 \times 10^{0}$
(c) $3 \times 10^{4}+7 \times 10^{2}+5 \times 10^{0}$
(d) $9 \times 10^{5}+2 \times 10^{2}+3 \times 10^{1}$
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Sign up nowExpress the following numbers in standard form:
(i) $5,00,00,000$
(ii) $70,00,000$
(iii) $3,18,65,00,000$
(iv) $3,90,878$
(v) 39087.8
(vi) 3908.78
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Sign up nowExpress the number appearing in the following statements in standard form.
(a) The distance between Earth and Moon is $384,000,000 \mathrm{~m}$.
(b) Speed of light in vacuum is $300,000,000 \mathrm{~m} / \mathrm{s}$.
(c) Diameter of the Earth is 1,27,56,000 m.
(d) Diameter of the Sun is $1,400,000,000 \mathrm{~m}$.
(e) In a galaxy there are on an average 100,000,000,000 stars.
(f) The universe is estimated to be about 12,000,000,000 years old.
(g) The distance of the Sun from the centre of the Milky Way Galaxy is estimated to be $300,000,000,000,000,000,000 \mathrm{~m}$.
(h) $60,230,000,000,000,000,000,000$ molecules are contained in a drop of water weighing $1.8 \mathrm{gm}$.
(i) The earth has 1,353,000,000 cubic $\mathrm{km}$ of sea water.
(j) The population of India was about 1,027,000,000 in March, 2001.
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Sign up nowExtra Questions - Exponents and Powers | NCERT | Mathematics | Class 7
If $(2)^{m+1} \times 2^{5} = 4^{-2}$, the value of $m$ is
(A) -7
(B) 7
(C) 10
(D) -10
Solution:
The correct option is **(D) -10**
Given equation is:
$$(2)^{m+1} \times 2^{5} = 4^{-2}$$
First, recognize that $4^{-2}$ can be written in terms of base $2$:
$$
4^{-2} = (2^2)^{-2}
$$
Using the property $\left(a^m\right)^n = a^{m \times n}$:
$$
(2^2)^{-2} = 2^{2 \times -2} = 2^{-4}
$$
Now, simplify the left-hand side (LHS) using the property $a^m \times a^n = a^{m+n}$:
$$
(2)^{m+1} \times 2^{5} = 2^{m+1+5} = 2^{m+6}
$$
Therefore, you now have:
$$
2^{m+6} = 2^{-4}
$$
Given that the bases are the same, set the exponents equal:
$$
m+6 = -4
$$
Solving for $m$:
$$
m = -4 - 6 = -10
$$
**Hence, $m = -10$**, which corresponds to option **(D)**.
Choose the most appropriate option to replace (?).
$$ 435 354 282 219 165 \text{ ? } $$
(A) 103
(B) 112
(C) 120
(D) $\mathbf{130}$
(E) None of these
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Sign up now"If $\log_{2}x * \log_{\frac{x}{64}}2 = \log_{\frac{x}{16}}2$, then the value of $x$ is:
(A) 2
(B) 4
(C) 16
(D) 8"
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Sign up nowWhich of the following hold good? If $n$ is a positive integer, then
A $n^{n}>1.3 \times 5 \times \ldots \times (2n-1)$
B $2.4 \times 6 \times \ldots \times 2n < (n+1)^{n}$
C $(n!)^{3} < n^{n} \left(\frac{n+1}{2}\right)^{2n}$
D $\left[1^{r}+2^{r}+3^{r}+\ldots+n^{r}\right]^{n} > n^{n} \times (n!)^{r}$
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