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In Class 9 Mathematics, the topic Exponents builds on Number Systems by showing how repeated multiplication is represented compactly using powers. Students learn to identify the base and exponent, apply the laws of exponents while multiplying and dividing powers, and work with zero and negative integral exponents. They practise simplifying numerical and algebraic expressions, compare powers, and use exponent notation accurately to express very large or very small numbers.
TOPIC PRACTICE
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25 questions
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Expert · Level 5View options
1
2
Any real number
4
Expert · Level 5View options
\(\frac{9}{4}\)
\(\frac{3}{2}\)
\(\frac{27}{4}\)
\(\frac{81}{16}\)
Expert · Level 5View options
\(\frac{7}{4}\)
2
\(\frac{5}{2}\)
3
Expert · Level 5View options
a^7
a^{12}
a^1
a^0
Expert · Level 5View options
x^7
x^{10}
x^3
x
Expert · Level 5View options
a^6
a^5
a^9
a^8
Expert · Level 5View options
2^7
2^{12}
2^1
2^0
Expert · Level 5View options
\(a^4\)
\(a^8\)
\(a^3\)
\(a^2\)
Expert · Level 5View options
\(x^3\)
\(x^{13}\)
\(x^2\)
\(x\)
Expert · Level 5View options
1
0
a
-1
Expert · Level 5View options
x^6
x^5
x^9
x^3
Expert · Level 5View options
\(2^6\)
\(2^5\)
\(2^8\)
\(2^4\)
Expert · Level 5View options
3^5
3^4
3^6
3^1
Expert · Level 5View options
\(2^4\)
\(2^{10}\)
\(2^3\)
\(2^1\)
Expert · Level 5View options
\(\frac{1}{a^2}\)
\(a^2\)
\(-a^2\)
\(\frac{1}{a}\)
Expert · Level 5View options
\(\frac{1}{x^3}\)
\(x^3\)
\(-x^3\)
1
Expert · Level 5View options
\(a^4\)
\(\frac{1}{a^4}\)
\(a^{-4}\)
1
Expert · Level 5View options
\(a^{-3}\)
\(a^3\)
\(a^{-2}\)
1
Expert · Level 5View options
\(2^{-4}\)
\(2^{4}\)
\(2^{-2}\)
1
Expert · Level 5View options
\(a^2b\)
\(a^3b^2\)
\(ab\)
1
Expert · Level 5View options
\(a^2b^2\)
\(ab^2\)
\(a^2b\)
\(a+b\)
Expert · Level 5View options
\(2^9\)
\(2^{10}\)
\(2^7\)
\(2^8\)
Expert · Level 5View options
4
2
8
16
Expert · Level 5View options
\(x^4 y^6\)
\(x^2 y^6\)
\(x^4 y^3\)
\(xy^5\)
Expert · Level 5View options
3
9
1
27
Question 1ExpertLevel 5
If (\frac{2^{x+3}}{2^{x-1}}=16), what is the value of (x)?
Correct answer: C
For division of powers with the same base, subtract the exponents: \(\frac{2^{x+3}}{2^{x-1}}=2^{(x+3)-(x-1)}=2^4=16\). Thus, the given equation is true for every real value of \(x\), so there is no single fixed value of \(x\). Therefore, option C is correct. Exam tip: use \(a^m/a^n=a^{m-n}\) for powers with the same non-zero base; here, \(x\) cancels out.
What is the value of \(\left(\frac{27}{8}\right)^{2/3}\)?
Correct answer: A
Using the fractional exponent rule \(a^{m/n}=(\sqrt[n]{a})^m\), we get \(\left(\frac{27}{8}\right)^{2/3}=\left(\sqrt[3]{\frac{27}{8}}\right)^2=\left(\frac{3}{2}\right)^2=\frac{9}{4}\). Therefore, option A is correct. Option B is only the cube root, \(\frac{3}{2}\), and does not include the required square. Exam tip: for an exponent \(m/n\), take the \(n\)th root and then raise the result to the power \(m\).
Express 4 and 64 as powers of 2: \(4^{x+1}=2^{2x+2}\) and \(64^{x-1}=2^{6x-6}\). Since the bases are equal, equate the exponents: \(2x+2=6x-6\). Thus, \(8=4x\), giving \(x=2\). Exam tip: When an exponential equation is reduced to the same base, equate the exponents.
According to laws of exponents what is (a^3 \times a^4)
Correct answer: A
When powers with the same base are multiplied, their exponents are added: \(a^m \times a^n = a^{m+n}\). Thus, \(a^3 \times a^4 = a^{3+4}=a^7\), so option A is correct. Option B incorrectly multiplies the exponents, which is not the rule for multiplication of like bases. Exam tip: add exponents for multiplication of like bases and subtract them for division.
When powers with the same base are multiplied, their exponents are added: x^5 × x^2 = x^{5+2} = x^7. Therefore, option A is correct. Option B incorrectly multiplies the exponents, whereas this rule requires their addition. Exam tip: for multiplication of like bases, keep the base unchanged and add the exponents.
For a power raised to another power, use the rule \((a^m)^n=a^{mn}\). Therefore, \((a^2)^3=a^{2\times3}=a^6\), so option A is correct. Option C is incorrect because it adds the exponents to get 9; exponents are multiplied in this situation. Exam tip: When a power is raised to another power, multiply the exponents.
When powers with the same base are multiplied, their exponents are added: 2^3 × 2^4 = 2^(3+4) = 2^7. Therefore, option A is correct. Option B incorrectly multiplies the exponents, which is not the rule for multiplying like bases. Exam tip: remember a^m × a^n = a^(m+n) for multiplication of powers with the same base.
When powers with the same non-zero base are divided, their exponents are subtracted: \(\frac{a^m}{a^n}=a^{m-n}\), where \(a\ne0\). Thus, \(\frac{a^6}{a^2}=a^{6-2}=a^4\). Exam tip: subtract exponents for division of like bases; do not add them.
For division of powers with the same non-zero base, subtract the exponents: \(\frac{x^m}{x^n}=x^{m-n}\), where \(x\neq0\). Thus, \(\frac{x^8}{x^5}=x^{8-5}=x^3\). Option B incorrectly adds the exponents, which is the rule for multiplication, not division. Exam tip: add exponents in multiplication and subtract them in division when the bases are the same.
For any non-zero number or variable, the exponent rule gives \(a^m \div a^m=a^{m-m}=a^0\). The left-hand side equals 1, so \(a^0=1\) for \(a\neq 0\). Therefore, option A is correct. Option C would apply when the exponent is 1, not 0. Exam tip: the zeroth power of every non-zero base is 1.
When a power is raised to another power, the exponents are multiplied: \((x^3)^2=x^{3\times2}=x^6\). Therefore, option A is correct. Remember that the exponents are multiplied, not added, so \(x^5\) is incorrect.
Use the power-of-a-power rule: \((a^m)^n=a^{mn}\). Therefore, \((2^2)^3=2^{2\times3}=2^6=64\), so option A is correct. Option C is incorrect because the exponents are not combined by adding or arbitrarily increasing them; they must be multiplied. Exam tip: when a power is raised to another power, multiply the exponents.
When powers with the same base are multiplied, their exponents are added: 3^4 × 3^1 = 3^(4+1) = 3^5. Therefore, option A is correct. Remember: add exponents for multiplication with the same base, but subtract them for division.
For division of powers with the same base, use \(\frac{a^m}{a^n}=a^{m-n}\). Thus, \(\frac{2^7}{2^3}=2^{7-3}=2^4\), so option A is correct. Option B results from adding the exponents, which is the rule for multiplication, not division. Exam tip: subtract the exponents when dividing powers with the same non-zero base.
The rule for a negative exponent is \(a^{-n}=\frac{1}{a^n}\), where \(a\neq0\). Therefore, \(a^{-2}=\frac{1}{a^2}\). Option B is the positive exponent form, while option D represents only \(a^{-1}\). Exam tip: for a negative exponent, take the reciprocal of the base and change the exponent to positive.
The negative-exponent rule is \(x^{-n}=\frac{1}{x^n}\), where \(x\neq0\). Therefore, \(x^{-3}=\frac{1}{x^3}\). Option B represents a positive exponent, while option C incorrectly adds a minus sign instead of taking the reciprocal. Exam tip: for a negative exponent, take the reciprocal of the base and make the exponent positive.
For \(a\neq 0\), the negative-exponent rule is \(a^{-n}=\frac{1}{a^n}\). Thus, \(a^{-4}=\frac{1}{a^4}\), so \(\frac{1}{a^{-4}}=\frac{1}{1/a^4}=a^4\). Therefore, option A is correct. Exam tip: first rewrite a negative exponent as a reciprocal, then simplify the resulting fraction.
Using the power-of-a-power rule, \((a^m)^n=a^{mn}\). Therefore, \((a^{-1})^3=a^{(-1)\times3}=a^{-3}\), provided \(a\ne0\). Option B ignores the negative exponent, while option C results from an incorrect calculation of the exponents. Exam tip: When a power is raised to another power, multiply the exponents.
Using the power-of-a-power rule \((a^m)^n=a^{mn}\), \((2^{-2})^2=2^{(-2)\times2}=2^{-4}=\frac{1}{16}\). Therefore, option A is correct. Option B is incorrect because it changes the negative exponent to a positive one. Exam tip: when a power is raised to another power, multiply the exponents.
For like bases in a quotient, subtract the exponent in the denominator from the exponent in the numerator: \(a^{3-1}b^{2-1}=a^2b\). Therefore, option A is correct. Option C is incorrect because it reduces both factors too far. Exam tip: use \(\frac{x^m}{x^n}=x^{m-n}\) for the same nonzero base; here \(a\neq0\) and \(b\neq0\) are assumed.
The power of a product follows the rule \((xy)^n=x^ny^n\). Therefore, \((ab)^2=a^2b^2\), so option A is correct. Options B and C apply the exponent to only one variable, while option D incorrectly changes multiplication into addition. Exam tip: when a product is raised to a power, apply that power to every factor.
Write 4 as \(2^2\). Then \(2^5 \times 4^2 = 2^5 \times (2^2)^2 = 2^5 \times 2^4 = 2^{5+4}=2^9\). Therefore, option A is correct. Option B results from adding the exponents incorrectly. Exam tip: when powers with the same base are multiplied, their exponents are added.
Rewrite 8 as a power of 2: 8 = 2^3, so 8^2 = (2^3)^2 = 2^6. Therefore, \(\frac{8^2}{2^4}=\frac{2^6}{2^4}=2^{6-4}=2^2=4\). Hence, option A is correct; option D results from considering only the denominator and ignoring the numerator. In exams, subtract the exponents when dividing powers with the same base.
Using the law \((ab)^n=a^n b^n\), we get \((x^2y^3)^2=(x^2)^2(y^3)^2=x^{2\times2}y^{3\times2}=x^4y^6\). Therefore, option A is correct. In option B, the exponent of \(x\) was not multiplied by 2, while option C leaves the exponent of \(y\) unchanged. Exam tip: when a product is raised to a power, apply that power to every factor and multiply the exponents.
When powers with the same base are multiplied, their exponents are added: \(3^{-1} \times 3^2 = 3^{-1+2} = 3^1 = 3\). Therefore, the correct answer is 3. Exam tip: do not add the bases; apply the law \(a^m \times a^n = a^{m+n}\).
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