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CBSE - Class 9 Mathematics Number Systems Worksheet
EXERCISE 1.5
Worksheet Answers
Solution:
We are tasked with evaluating the following exponential expression with a rational exponent:
$16^{\frac{3}{4}}$
To solve this, we must apply the fundamental laws of exponents. There are two mathematically equivalent pathways to arrive at the solution: utilizing prime factorization or converting the expression into radical notation.
The following diagram illustrates the sequential transformation of the base $16$ through the operations dictated by the fractional exponent $\frac{3}{4}$.
Final Solution: $16^{\frac{3}{4}} = 8$
Solution:
We are required to evaluate the exponential expression:
$32^{\frac{2}{5}}$
This expression features a base of $32$ raised to a rational (fractional) exponent $\frac{2}{5}$. To simplify this, we must decompose the base into its fundamental prime factors.
We begin by finding the prime factorization of the integer $32$. By continuously dividing by the smallest prime number ($2$), we can express $32$ as a power of $2$.
Thus, the prime factorization is:
$32 = 2 \times 2 \times 2 \times 2 \times 2 = 2^5$
Substitute the prime factorized form ($2^5$) back into the original expression in place of $32$:
$ (2^5)^{\frac{2}{5}} $
To simplify this, we apply the Power of a Power Property of exponents [Per the fundamental law of exponents which states that $(a^m)^n = a^{m \cdot n}$]. This allows us to multiply the inner exponent by the outer rational exponent:
$ 2^{5 \times \frac{2}{5}} $
Now, perform the multiplication in the exponent. The integer $5$ in the numerator cancels out perfectly with the denominator $5$ of the fraction:
$ 5 \times \frac{2}{5} = \frac{5 \times 2}{5} = \frac{10}{5} = 2 $
Substituting the simplified exponent back onto the base of $2$ yields:
$ 2^2 $
Evaluate the simplified exponential expression by squaring the base:
$ 2^2 = 2 \times 2 = 4 $
Final Solution: 4
Solution:
We are tasked with evaluating the following exponential expression:
$64^{\frac{1}{2}}$
[Per the Definition of Rational Exponents], an expression of the form $a^{\frac{m}{n}}$ is mathematically equivalent to the $n$-th root of $a$ raised to the $m$-th power, expressed as $\sqrt[n]{a^m}$.
In our specific case, the exponent is $\frac{1}{2}$. This indicates the principal square root of the base. Therefore:
$64^{\frac{1}{2}} = \sqrt[2]{64^1} = \sqrt{64}$
To evaluate this expression algebraically without relying on memorized perfect squares, we first determine the prime factorization of the base, $64$.
Thus, the prime factorization is $64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6$.
Substituting this back into our original expression yields:
$(2^6)^{\frac{1}{2}}$
[Per the Laws of Exponents], specifically the Power of a Power property which states that $(a^m)^n = a^{m \cdot n}$, we multiply the inner exponent by the outer fractional exponent:
$(2^6)^{\frac{1}{2}} = 2^{6 \times \frac{1}{2}}$
$2^{6 \times \frac{1}{2}} = 2^3$
Evaluating $2^3$ gives:
$2 \times 2 \times 2 = 8$
Alternatively, recognizing that $64$ is a perfect square provides a more direct algebraic route. We know that $8 \times 8 = 64$, meaning $64 = 8^2$.
Substituting $8^2$ into the expression:
$(8^2)^{\frac{1}{2}}$
Applying the same Power of a Power property $(a^m)^n = a^{m \cdot n}$:
$8^{2 \times \frac{1}{2}} = 8^1 = 8$
Geometrically, finding the square root of $64$ (i.e., $64^{\frac{1}{2}}$) is equivalent to finding the side length of a square whose total area is $64$ square units. The diagram below illustrates an $8 \times 8$ grid, visually proving that a side length of $8$ yields an area of $64$.
Through both prime factorization and perfect square identification, governed by the fundamental laws of exponents, the expression simplifies perfectly to the integer $8$.
Final Solution: 8
Solution:
We are tasked with simplifying the following mathematical expression involving fractional exponents:
$2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}$
By analyzing the expression, we identify the following components:
To simplify the multiplication of two exponential terms with the same base, we apply the Product Rule for Exponents. [Per the fundamental laws of exponents for real numbers], when multiplying two expressions with identical bases, their exponents are added together while the base remains unchanged.
The algebraic theorem is stated as:
$a^m \cdot a^n = a^{m+n}$
Substituting our identified variables into this theorem yields:
$2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}} = 2^{\left(\frac{2}{3} + \frac{1}{5}\right)}$
We must now evaluate the sum of the fractional exponents: $\frac{2}{3} + \frac{1}{5}$.
To add fractions with different denominators, we must first determine their Least Common Multiple (LCM). The denominators are $3$ and $5$. Since both are prime numbers, their LCM is simply their product:
$\text{LCM}(3, 5) = 3 \times 5 = 15$
Next, we convert each fraction into an equivalent fraction with a denominator of $15$:
Now, substitute the equivalent fractions back into the exponent sum and evaluate:
$\frac{2}{3} + \frac{1}{5} = \frac{10}{15} + \frac{3}{15} = \frac{10 + 3}{15} = \frac{13}{15}$
Replacing this sum back into our exponential expression gives the final simplified form:
$2^{\left(\frac{2}{3} + \frac{1}{5}\right)} = 2^{\frac{13}{15}}$
Final Solution: The simplified form of the expression is $2^{\frac{13}{15}}$.
Solution:
We are tasked with simplifying the following exponential expression:
$ E = \left(\frac{1}{3^3}\right)^7 $
To simplify this expression rigorously, we will apply the fundamental Laws of Exponents. There are two primary analytical pathways to arrive at the solution. Both methods are detailed below to ensure comprehensive understanding.
Step 1: Transform the Fractional Base
[Per the Reciprocal Law of Exponents], any non-zero base raised to a positive power in the denominator can be expressed as a negative exponent in the numerator. The governing axiom is:
$ \frac{1}{a^n} = a^{-n} $
Applying this theorem to the inner fraction of our expression:
$ \frac{1}{3^3} = 3^{-3} $
Step 2: Substitute and Apply the Power of a Power Rule
Substitute the transformed base back into the original expression:
$ E = \left(3^{-3}\right)^7 $
[Per the Power of a Power Property], when an exponential term is raised to another exponent, the exponents are multiplied. The governing axiom is:
$ (a^m)^n = a^{m \cdot n} $
Applying this property to our expression yields:
$ E = 3^{-3 \cdot 7} $
Step 3: Arithmetic Simplification
Perform the multiplication in the exponent:
$ -3 \cdot 7 = -21 $
$ E = 3^{-21} $
Step 1: Distribute the Exponent
[Per the Power of a Quotient Rule], an exponent applied to a fraction is distributed to both the numerator and the denominator. The governing axiom is:
$ \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} $
Applying this to our expression:
$ E = \frac{1^7}{(3^3)^7} $
Step 2: Simplify Numerator and Denominator
Since $1$ raised to any finite power is $1$ ($1^7 = 1$), the numerator simplifies immediately. For the denominator, we again apply the Power of a Power Property $(a^m)^n = a^{m \cdot n}$:
$ E = \frac{1}{3^{3 \cdot 7}} $
$ E = \frac{1}{3^{21}} $
Step 3: Convert to a Negative Exponent
Using the Reciprocal Law of Exponents $\left(\frac{1}{a^n} = a^{-n}\right)$ to express the final answer without a fraction:
$ E = 3^{-21} $
The following diagram illustrates the sequential application of the exponent rules used in Method 1.
Final Solution: The simplified expression is $3^{-21}$ (which can also be written as $\frac{1}{3^{21}}$).
Solution:
We are tasked with simplifying the following exponential expression:
$ \frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}} $
Here, we identify the components of the expression:
To simplify the division of two exponential terms with the same base, we apply the Quotient Law of Exponents. [Per the fundamental laws of algebra for real numbers], when dividing like bases, we subtract the exponent of the denominator from the exponent of the numerator.
The formal theorem states:
$ \frac{a^m}{a^n} = a^{m - n} $
Condition: $a > 0$ and $a \neq 1$. Since our base is $11$, this condition is perfectly satisfied.
Substituting our specific values ($a = 11$, $m = \frac{1}{2}$, $n = \frac{1}{4}$) into the formula, we get:
$ \frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}} = 11^{\left(\frac{1}{2} - \frac{1}{4}\right)} $
We must now perform the arithmetic operation on the exponents: $\frac{1}{2} - \frac{1}{4}$.
[By the rules of rational number subtraction], we must first find a common denominator for the fractions. The Least Common Multiple (LCM) of the denominators $2$ and $4$ is $4$.
Now, substitute the equivalent fraction back into the exponent expression:
$ \frac{2}{4} - \frac{1}{4} = \frac{2 - 1}{4} = \frac{1}{4} $
Replacing the simplified exponent back onto our base of $11$, we obtain the final simplified form:
$ 11^{\frac{1}{4}} $
This can also be written in radical notation as the fourth root of $11$, or $\sqrt[4]{11}$, but standard exponential simplification leaves it in fractional exponent form unless otherwise specified.
Final Solution: $ 11^{\frac{1}{4}} $
Solution:
We are tasked with evaluating the exponential expression $125^{\frac{1}{3}}$.
[Per the definition of rational exponents, an expression of the form $a^{\frac{1}{n}}$ represents the $n$-th principal root of the real number $a$. Therefore, evaluating $125^{\frac{1}{3}}$ is algebraically equivalent to finding the principal cube root of $125$, denoted as $\sqrt[3]{125}$].
To simplify the expression systematically, we first determine the prime factorization of the base, $125$. We divide $125$ by its smallest prime factor, which is $5$:
Thus, the prime factorization is $125 = 5 \times 5 \times 5 = 5^3$.
[By the Fundamental Theorem of Arithmetic, every integer greater than $1$ can be represented uniquely as a product of prime numbers, up to the order of the factors].
Geometrically, finding the cube root of $125$ is equivalent to finding the side length $s$ of a perfect cube whose total volume is $125$ cubic units. The relationship is defined by the volume formula $V = s^3$.
Substitute the prime factored form of the base back into the given expression:
$125^{\frac{1}{3}} = (5^3)^{\frac{1}{3}}$
To simplify $(5^3)^{\frac{1}{3}}$, we apply the Power of a Power Property.
[The Power of a Power Property states that $(a^m)^n = a^{m \cdot n}$. This law dictates that when an exponential term is raised to another power, the exponents must be multiplied].
Applying this theorem:
$(5^3)^{\frac{1}{3}} = 5^{3 \cdot \frac{1}{3}}$
Perform the arithmetic multiplication in the exponent:
$3 \cdot \frac{1}{3} = \frac{3}{3} = 1$
Thus, the expression simplifies to:
$5^1$
[By the identity property of exponents, any non-zero real number raised to the power of $1$ is the number itself, meaning $a^1 = a$].
$5^1 = 5$
Final Solution: 5
Solution:
We are tasked with simplifying the following mathematical expression involving rational exponents:
$7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}$
Upon analyzing the expression, we identify two distinct bases ($a = 7$ and $b = 8$) raised to an identical rational power ($m = \frac{1}{2}$).
To simplify the product of two different bases raised to the same exponent, we apply the Multiplicative Law of Exponents for Identical Powers. [Per the fundamental axioms of real number exponents], this law states:
$a^m \cdot b^m = (a \cdot b)^m$
This theorem holds true for any positive real numbers $a$ and $b$, and any rational number $m$.
By substituting our specific values into the established theorem ($a = 7$, $b = 8$, and $m = \frac{1}{2}$), we consolidate the bases inside a single parenthesis:
$7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}} = (7 \cdot 8)^{\frac{1}{2}}$
Next, we perform the arithmetic multiplication within the parentheses:
$7 \cdot 8 = 56$
Substituting this product back into the expression yields:
$56^{\frac{1}{2}}$
While $56^{\frac{1}{2}}$ is the simplified exponential form, it is mathematically rigorous to understand its equivalent radical form. [By the definition of rational exponents], an exponent of $\frac{1}{2}$ denotes the principal square root of the base:
$x^{\frac{1}{2}} = \sqrt{x}$
Therefore, our expression can be written as:
$\sqrt{56}$
To ensure the radical is in its simplest form, we extract the largest perfect square factor from the radicand (56). We perform prime factorization:
Applying the product property of radicals ($\sqrt{x \cdot y} = \sqrt{x} \cdot \sqrt{y}$):
$\sqrt{56} = \sqrt{4 \cdot 14} = \sqrt{4} \cdot \sqrt{14} = 2\sqrt{14}$
Final Solution: The simplified expression is $56^{\frac{1}{2}}$ (which can also be rigorously expressed in radical form as $2\sqrt{14}$).
Solution:
We are tasked with evaluating the exponential expression $125^{-\frac{1}{3}}$. To solve this, we will systematically apply the fundamental laws of exponents and prime factorization.
The base of the given expression is $125$. To simplify expressions involving fractional exponents, the standard procedure is to express the base as a power of its prime factors [Per the Fundamental Theorem of Arithmetic].
Dividing $125$ by its smallest prime factor ($5$):
Thus, the prime factorization is:
$125 = 5 \times 5 \times 5 = 5^3$
We substitute the prime factored form of the base back into the original expression.
$125^{-\frac{1}{3}} = (5^3)^{-\frac{1}{3}}$
[Per the Laws of Exponents], specifically the Power of a Power Property, we know that for any non-zero real number $a$ and rational numbers $m$ and $n$:
$(a^m)^n = a^{m \cdot n}$
Applying this theorem to our expression, we multiply the inner integer exponent ($3$) by the outer fractional exponent ($-\frac{1}{3}$):
$(5^3)^{-\frac{1}{3}} = 5^{3 \times \left(-\frac{1}{3}\right)}$
Perform the multiplication in the exponent:
$3 \times \left(-\frac{1}{3}\right) = -\frac{3}{3} = -1$
Substituting this back yields:
$5^{-1}$
[Per the Negative Exponent Rule], a negative exponent indicates the reciprocal of the base raised to the corresponding positive exponent. Mathematically, for any non-zero real number $a$:
$a^{-n} = \frac{1}{a^n}$
Applying this to our simplified expression:
$5^{-1} = \frac{1}{5^1} = \frac{1}{5}$
Final Solution: The value of $125^{-\frac{1}{3}}$ is $\frac{1}{5}$.
Solution:
We are tasked with evaluating the following exponential expression:
$32^{\frac{1}{5}}$
[Per the definition of rational exponents, an expression of the form $a^{\frac{1}{n}}$ represents the $n$-th principal root of the base $a$, denoted algebraically as $\sqrt[n]{a}$. Therefore, evaluating $32^{\frac{1}{5}}$ is mathematically equivalent to finding the 5th root of $32$.]
To systematically simplify expressions involving fractional exponents, we first decompose the base ($32$) into its prime factors. This allows us to express the base as an exponential term, which can then be manipulated using the laws of exponents.
Dividing $32$ successively by the smallest prime number ($2$):
Thus, the prime factorization is $32 = 2 \times 2 \times 2 \times 2 \times 2 = 2^5$.
We substitute the prime factored form of the base back into the original expression:
$32^{\frac{1}{5}} = (2^5)^{\frac{1}{5}}$
[Per the Power of a Power Property of Exponents], which states that $(a^m)^n = a^{m \times n}$, we multiply the inner exponent ($5$) by the outer fractional exponent ($\frac{1}{5}$).
$(2^5)^{\frac{1}{5}} = 2^{5 \times \frac{1}{5}}$
We now perform the multiplication in the exponent:
$5 \times \frac{1}{5} = \frac{5}{5} = 1$
Substituting this simplified exponent back onto the base $2$ yields:
$2^1 = 2$
Final Solution: 2
Solution:
We are tasked with evaluating the exponential expression with a rational exponent:
$9^{\frac{3}{2}}$
To solve this, we can utilize the fundamental laws of exponents. There are two mathematically rigorous pathways to arrive at the solution: Prime Factorization (Power of a Power Rule) and Radical Conversion. Both methods are detailed below to ensure comprehensive understanding.
The base of the expression is $9$. We first express $9$ in its prime factorized form. [Per the Fundamental Theorem of Arithmetic, every integer greater than $1$ can be represented uniquely as a product of prime numbers].
$9 = 3 \times 3 = 3^2$
Substitute the prime factorized form back into the original expression:
$9^{\frac{3}{2}} = (3^2)^{\frac{3}{2}}$
Next, we apply the Power of a Power Property of exponents, which states that for any non-zero base $a$ and rational exponents $m$ and $n$:
$(a^m)^n = a^{m \cdot n}$
Applying this property to our expression:
$(3^2)^{\frac{3}{2}} = 3^{2 \times \frac{3}{2}}$
Multiply the exponents. The integer $2$ in the numerator cancels out the denominator $2$ in the fractional exponent:
$2 \times \frac{3}{2} = \frac{2 \times 3}{2} = 3$
Thus, the expression simplifies to a base raised to an integer power:
$3^3$
Expand the cubic expression to find the final numerical value:
$3^3 = 3 \times 3 \times 3 = 27$
[Per the definition of rational exponents, an expression of the form $a^{\frac{m}{n}}$ can be rewritten as the $n$-th root of $a$ raised to the $m$-th power]:
$a^{\frac{m}{n}} = (\sqrt[n]{a})^m$
Applying this theorem to $9^{\frac{3}{2}}$, where $a = 9$, $m = 3$, and $n = 2$:
$9^{\frac{3}{2}} = (\sqrt[2]{9})^3$
Evaluate the principal square root of $9$:
$\sqrt{9} = 3$
Substitute this back into the expression:
$(3)^3 = 27$
Both analytical pathways yield the exact same scalar quantity, verifying the integrity of the result.
The following diagram illustrates the structural equivalence of both methods used to evaluate the rational exponent.
Final Solution: $9^{\frac{3}{2}} = 27$
