OP Malhotra Class 9 Maths Solutions Chapter 6 Indices Exponents Chapter Test

S Chand Class 9 ICSE Maths Solutions Chapter 6 Indices/Exponents Chapter Test

Question 1. Determine whether each equation is true or false. Change the right side of the equation to make a true equation.
(i) \( (2a)^{-3} = \frac{2}{a^3} \)
(ii) \( \left(\left(a^{-1}\right)^{-1}\right)^{-1}=\frac{1}{a} \)
(iii) \( (2 + 3)^{-1} = 2^{-1} + 3^{-1} \)
(iv) \( (3x – 1)^0 = (1 – 3x)^0 \) Why is the condition \( x \neq \frac { 1 }{ 3 } \) given in part (iv) above?
Answer:
(i) The given equation is \( (2a)^{-3} = \frac{2}{a^3} \). When we simplify the left side, \( (2a)^{-3} \) means \( 1 \) divided by \( (2a)^3 \). This becomes \( 1 \) divided by \( 8a^3 \). So, the equation is false because \( \frac{1}{8a^3} \) is not equal to \( \frac{2}{a^3} \). To make it true, the right side should be \( \frac{1}{8a^3} \). Raising a term to a negative power means taking its reciprocal.
(ii) The given equation is \( \left(\left(a^{-1}\right)^{-1}\right)^{-1}=\frac{1}{a} \). We use the power rule \( (x^m)^n = x^{m \times n} \). Applying this, \( a^{-1} \) raised to \( (-1) \) gives \( a^{(-1) \times (-1)} = a^1 \). Then \( a^1 \) raised to \( (-1) \) gives \( a^{1 \times (-1)} = a^{-1} \). \( a^{-1} \) is equal to \( \frac{1}{a} \). So, the equation is true. Negative exponents indicate reciprocals.
(iii) The given equation is \( (2 + 3)^{-1} = 2^{-1} + 3^{-1} \). First, simplify the left side: \( (2+3)^{-1} = 5^{-1} = \frac{1}{5} \). Next, simplify the right side: \( 2^{-1} + 3^{-1} = \frac{1}{2} + \frac{1}{3} \). To add these fractions, find a common denominator, which is 6. So, \( \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \). Since \( \frac{1}{5} \neq \frac{5}{6} \), the equation is false. To make it true, the right side should be \( \frac{1}{5} \). The sum of reciprocals is not the reciprocal of the sum.
(iv) The given equation is \( (3x - 1)^0 = (1 - 3x)^0 \). Any non-zero number raised to the power of zero is 1. So, \( (3x - 1)^0 = 1 \) and \( (1 - 3x)^0 = 1 \). Since \( 1 = 1 \), the equation is true.
Regarding the condition \( x \neq \frac{1}{3} \): This condition is given because if \( x \) were equal to \( \frac{1}{3} \), then \( (3x-1) \) would become \( (3 \times \frac{1}{3} - 1) = (1-1) = 0 \). In mathematics, \( 0^0 \) is an undefined form. Therefore, to ensure that \( (3x-1)^0 \) and \( (1-3x)^0 \) are both equal to 1, we must have \( (3x-1) \neq 0 \), which means \( x \neq \frac{1}{3} \). It ensures the base is not zero before applying the zero exponent rule.
In simple words: We check each equation. For those that are false, we correct the right side. We also explain why certain conditions, like \( x \neq \frac{1}{3} \), are necessary in some equations.

🎯 Exam Tip: Remember that \( a^{-n} = \frac{1}{a^n} \) and \( a^0 = 1 \) for any non-zero \( a \). Always apply the rules of exponents carefully to each term.

Simplify

Question 2. \( \frac{\left(5 x^3 y^{-3} z\right)^{-2}}{y^4 z^{-2}} \)
Answer: We need to simplify the given expression \( \frac{\left(5 x^3 y^{-3} z\right)^{-2}}{y^4 z^{-2}} \).
First, apply the power of -2 to each term in the numerator: \( (5^{-2}) (x^{3 \times -2}) (y^{-3 \times -2}) (z^{-2}) \). This simplifies to \( 5^{-2} x^{-6} y^6 z^{-2} \).
Next, rewrite \( 5^{-2} \) as \( \frac{1}{5^2} = \frac{1}{25} \).
So, the numerator becomes \( \frac{1}{25} x^{-6} y^6 z^{-2} \).
Now the expression is \( \frac{\frac{1}{25} x^{-6} y^6 z^{-2}}{y^4 z^{-2}} \).
Combine the terms with the same base by subtracting the exponents:
For \( y \): \( y^{6-4} = y^2 \).
For \( z \): \( z^{-2 - (-2)} = z^{-2+2} = z^0 = 1 \).
For \( x \): \( x^{-6} \) remains \( x^{-6} \).
So, we get \( \frac{1}{25} x^{-6} y^2 \times 1 \).
Finally, \( x^{-6} \) can be written as \( \frac{1}{x^6} \).
Thus, the simplified expression is \( \frac{y^2}{25x^6} \). Remember to apply negative exponents carefully.
In simple words: To simplify this, first, we apply the outer exponent to everything inside the bracket in the top part. Then, we combine similar letters (variables) by adding or subtracting their powers, according to the rules of exponents.

🎯 Exam Tip: When dealing with negative exponents like \( (ab)^{-n} \), apply the negative exponent to each term individually, i.e., \( a^{-n}b^{-n} \). Also, remember that \( x^0 = 1 \).

Question 3. \( \left(\frac{8 a^3 b^{-4}}{64 a^{-9} b^2}\right)^{\frac{2}{3}} \)
Answer: We need to simplify the expression \( \left(\frac{8 a^3 b^{-4}}{64 a^{-9} b^2}\right)^{\frac{2}{3}} \).
First, simplify inside the bracket. Divide 8 by 64 to get \( \frac{1}{8} \).
For \( a \) terms, subtract exponents: \( a^{3 - (-9)} = a^{3+9} = a^{12} \).
For \( b \) terms, subtract exponents: \( b^{-4 - 2} = b^{-6} \).
So the expression inside the bracket becomes \( \left(\frac{1}{8} a^{12} b^{-6}\right) \).
Now, raise each part to the power of \( \frac{2}{3} \).
\( \left(\frac{1}{8}\right)^{\frac{2}{3}} = (2^{-3})^{\frac{2}{3}} = 2^{-3 \times \frac{2}{3}} = 2^{-2} = \frac{1}{2^2} = \frac{1}{4} \).
\( (a^{12})^{\frac{2}{3}} = a^{12 \times \frac{2}{3}} = a^8 \).
\( (b^{-6})^{\frac{2}{3}} = b^{-6 \times \frac{2}{3}} = b^{-4} \).
Combine these simplified parts: \( \frac{1}{4} \times a^8 \times b^{-4} \).
Since \( b^{-4} = \frac{1}{b^4} \), the final simplified form is \( \frac{a^8}{4b^4} \). Remember that \( \sqrt[3]{8} = 2 \).
In simple words: First, simplify the numbers and variables inside the bracket by dividing and combining powers. Then, apply the outer power to each part of the simplified expression.

🎯 Exam Tip: Remember to simplify the numerical coefficients and variable terms separately inside the bracket before applying the outer exponent. When dividing terms with the same base, subtract the exponents.

Question 4. \( – \sqrt[4]{16 a^4b^8} \)
Answer: We need to simplify the expression \( – \sqrt[4]{16 a^4b^8} \).
The \( 4^{th} \) root means raising to the power of \( \frac{1}{4} \).
So, \( \sqrt[4]{16} = \sqrt[4]{2^4} = 2 \).
\( \sqrt[4]{a^4} = a^{4 \times \frac{1}{4}} = a^1 = a \).
\( \sqrt[4]{b^8} = b^{8 \times \frac{1}{4}} = b^2 \).
Now, multiply these terms together, keeping the negative sign outside the root.
So, \( – (2 \times a \times b^2) = – 2ab^2 \). Taking the nth root of a number is the same as raising it to the power of 1/n.
In simple words: We find the fourth root of each part inside the root symbol, like finding what number multiplied by itself four times gives that value. Then, we multiply these results together and keep the negative sign from the start.

🎯 Exam Tip: When simplifying roots of products like \( \sqrt[n]{abc} \), you can write it as \( \sqrt[n]{a} \times \sqrt[n]{b} \times \sqrt[n]{c} \). Remember to deal with the overall negative sign separately if it's outside the root.

Question 5. \( \left(\frac{y^{\frac{2}{3}} \cdot y^{-\frac{5}{6}}}{y^{\frac{1}{9}}}\right)^9 \)
Answer: To simplify \( \left(\frac{y^{\frac{2}{3}} \cdot y^{-\frac{5}{6}}}{y^{\frac{1}{9}}}\right)^9 \), we first simplify the expression inside the bracket.
When multiplying terms with the same base, we add their exponents: \( y^{\frac{2}{3} + (-\frac{5}{6})} = y^{\frac{4}{6} - \frac{5}{6}} = y^{-\frac{1}{6}} \).
So the numerator becomes \( y^{-\frac{1}{6}} \).
Now the expression inside the bracket is \( \frac{y^{-\frac{1}{6}}}{y^{\frac{1}{9}}} \).
When dividing terms with the same base, we subtract the exponent of the denominator from the exponent of the numerator: \( y^{-\frac{1}{6} - \frac{1}{9}} \).
Find a common denominator for 6 and 9, which is 18.
\( -\frac{1}{6} = -\frac{3}{18} \) and \( -\frac{1}{9} = -\frac{2}{18} \).
So, the exponent becomes \( -\frac{3}{18} - \frac{2}{18} = -\frac{5}{18} \).
The expression inside the bracket is \( y^{-\frac{5}{18}} \).
Finally, raise this to the power of 9: \( (y^{-\frac{5}{18}})^9 = y^{-\frac{5}{18} \times 9} = y^{-\frac{5}{2}} \).
This can also be written as \( \frac{1}{y^{\frac{5}{2}}} \). Combining exponents involves careful fraction arithmetic.
In simple words: First, simplify the powers of 'y' inside the bracket by adding exponents for multiplication and subtracting for division. Then, multiply this final exponent by the power outside the bracket.

🎯 Exam Tip: Always simplify inside the brackets first. When dealing with fractions as exponents, find a common denominator before adding or subtracting them.

Question 6. \( \sqrt[3]{\sqrt{x^6}} \)
Answer: We need to simplify the expression \( \sqrt[3]{\sqrt{x^6}} \).
First, consider the inner square root: \( \sqrt{x^6} \). A square root is the same as raising to the power of \( \frac{1}{2} \). So, \( (x^6)^{\frac{1}{2}} = x^{6 \times \frac{1}{2}} = x^3 \).
Now, take the cube root of this result: \( \sqrt[3]{x^3} \). A cube root is the same as raising to the power of \( \frac{1}{3} \). So, \( (x^3)^{\frac{1}{3}} = x^{3 \times \frac{1}{3}} = x^1 = x \).
Therefore, the simplified expression is \( x \). Roots can be written as fractional exponents for easier calculation.
In simple words: We change the roots into fractional powers. Then, we multiply these powers together and apply them to 'x' to get the simplest form.

🎯 Exam Tip: Remember that \( \sqrt[n]{a} = a^{\frac{1}{n}} \) and \( (a^m)^n = a^{mn} \). Convert roots to fractional exponents to simplify expressions with multiple roots easily.

Question 7. Which of the following is (are) equivalent to \( 16^{-\frac{1}{2}} \)?
(a) - 8
(b) \( \frac { 1 }{ 4 } \)
(c) – 4
(d) \( 4^{-1} \)
Answer: (b) \( \frac { 1 }{ 4 } \) and (d) \( 4^{-1} \)
In simple words: We need to simplify \( 16^{-\frac{1}{2}} \) first, which means finding the square root of 16 and then taking its reciprocal. This gives \( \frac{1}{4} \). Then we check which of the options also equals \( \frac{1}{4} \).

🎯 Exam Tip: When simplifying negative fractional exponents, break it down into reciprocal and root operations. \( a^{-\frac{m}{n}} = \frac{1}{\sqrt[n]{a^m}} \).

Question 8. Which of the following is undefined?
(a) – \( 25^{\frac { 1 }{ 2 }} \)
(b) \( 25^{\frac { 1 }{ 2 }} \)
(c) – \( 25^{-\frac { 1 }{ 2 }} \)
(d) \( (-25)^{\frac { 1 }{ 2 }} \)
Answer: (d) \( (-25)^{\frac { 1 }{ 2 }} \)
We need to find which of the given expressions is undefined in the real number system.
Let's evaluate each option:
(a) \( – 25^{\frac{1}{2}} = – \sqrt{25} = – 5 \). This is a defined number.
(b) \( 25^{\frac{1}{2}} = \sqrt{25} = 5 \). This is a defined number.
(c) \( – 25^{-\frac{1}{2}} = – \frac{1}{25^{\frac{1}{2}}} = – \frac{1}{\sqrt{25}} = – \frac{1}{5} \). This is a defined number.
(d) \( (-25)^{\frac{1}{2}} = \sqrt{-25} \). The square root of a negative number is not a real number. It involves imaginary numbers, but in the context of typical school math, it is considered undefined in the real number system.
Therefore, option (d) is undefined. Remember that you cannot take an even root of a negative number and get a real result.
In simple words: We check each option to see if its value exists as a normal number. The one that requires finding the square root of a negative number is undefined in real numbers.

🎯 Exam Tip: An even root (like square root, fourth root) of a negative number is undefined in the set of real numbers. An odd root (like cube root) of a negative number is defined.

Question 9. True or False?
(a) \( \frac{a^{4 n}}{a^n}=a^4 \)
(b) \( \frac{1}{a^{m-n}}=a^{n-m} \)
(c) \( a^{-n}.a^n = 1 \)
(d) \( \frac{a^n}{b^m}=\left(\frac{a}{b}\right)^{n-m} \)
Answer:
(a) The statement is \( \frac{a^{4n}}{a^n} = a^4 \). When dividing powers with the same base, we subtract the exponents. So, \( \frac{a^{4n}}{a^n} = a^{4n-n} = a^{3n} \). Since \( a^{3n} \) is not always equal to \( a^4 \) (unless \( 3n=4 \)), the statement is false.
(b) The statement is \( \frac{1}{a^{m-n}} = a^{n-m} \). A term \( \frac{1}{x^p} \) can be written as \( x^{-p} \). So, \( \frac{1}{a^{m-n}} \) becomes \( a^{-(m-n)} \). Distributing the negative sign gives \( a^{-m+n} \), which is the same as \( a^{n-m} \). Thus, the statement is true.
(c) The statement is \( a^{-n} \cdot a^n = 1 \). When multiplying powers with the same base, we add the exponents. So, \( a^{-n} \cdot a^n = a^{-n+n} = a^0 \). Any non-zero number raised to the power of zero is 1. So, \( a^0 = 1 \). Therefore, the statement is true. This is a fundamental rule of exponents.
(d) The statement is \( \frac{a^n}{b^m} = \left(\frac{a}{b}\right)^{n-m} \). The right side can be expanded as \( \frac{a^{n-m}}{b^{n-m}} \). For the original statement to be true, \( n \) must be equal to \( m \) for the exponents of \( a \) and \( b \) to match, which is not generally the case. For example, \( \frac{a^2}{b^3} \) is not equal to \( \left(\frac{a}{b}\right)^{2-3} = \left(\frac{a}{b}\right)^{-1} = \frac{b}{a} \). So, the statement is false. The powers of numerator and denominator are different here.
In simple words: We check each given rule about powers. We see if it holds true by using the basic rules of exponents for multiplication, division, and negative powers.

🎯 Exam Tip: Master the basic rules of exponents: \( a^m \times a^n = a^{m+n} \), \( \frac{a^m}{a^n} = a^{m-n} \), \( (a^m)^n = a^{mn} \), \( a^0=1 \) (for \( a \neq 0 \)), and \( a^{-n}=\frac{1}{a^n} \). Incorrect application of these rules is a common mistake.

Question 10.
(i) Solve : \( (- 4.8)^k = 1 \)
(ii) \( \sqrt[3]{\sqrt{0.000064}} \) is equal to
Answer:
(i) We need to solve \( (-4.8)^k = 1 \).
We know that any non-zero number raised to the power of zero is equal to 1. So, we can write 1 as \( (-4.8)^0 \).
This means \( (-4.8)^k = (-4.8)^0 \).
Since the bases are the same, the exponents must be equal. Therefore, \( k = 0 \). The zero exponent rule is key here.
(ii) We need to find the value of \( \sqrt[3]{\sqrt{0.000064}} \).
First, convert the roots into fractional exponents: \( \sqrt{\text{number}} \) is \( (\text{number})^{\frac{1}{2}} \), and \( \sqrt[3]{\text{number}} \) is \( (\text{number})^{\frac{1}{3}} \).
So the expression becomes \( ((0.000064)^{\frac{1}{2}})^{\frac{1}{3}} \).
When a power is raised to another power, we multiply the exponents: \( (0.000064)^{\frac{1}{2} \times \frac{1}{3}} = (0.000064)^{\frac{1}{6}} \).
Now, we need to find what number, when multiplied by itself six times, gives 0.000064.
We know that \( 2^6 = 64 \). So, \( (0.2)^6 = 0.2 \times 0.2 \times 0.2 \times 0.2 \times 0.2 \times 0.2 = 0.000064 \).
Substitute this back into the expression: \( ((0.2)^6)^{\frac{1}{6}} \).
Again, multiply the exponents: \( (0.2)^{6 \times \frac{1}{6}} = (0.2)^1 = 0.2 \).
Therefore, the value is \( 0.2 \). Understanding how decimal places work with powers is very helpful here.
In simple words: For part (i), we use the rule that any number to the power of zero is one. For part (ii), we change the roots into fractional powers, multiply them, and then find the base number that fits.

🎯 Exam Tip: For equations like \( a^x = 1 \), remember that \( x \) is usually 0 unless \( a \) itself is 1 or -1 (with certain conditions). For roots of decimals, try to express the decimal as a power of a smaller number.