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Q1(i):
Find :
(i) $64^{\frac{1}{2}}$
Solution :
Initial Setup & Given Expression
We are tasked with evaluating the following exponential expression:
$64^{\frac{1}{2}}$
Step 1: Theoretical Foundation of Fractional Exponents
[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:
$64^{\frac{1}{2}} = \sqrt[2]{64^1} = \sqrt{64}$
Step 2: Method 1 - Prime Factorization (Rigorous Approach)
To evaluate this expression algebraically without relying on memorized perfect squares, we first determine the prime factorization of the base, $64$.
- $64 \div 2 = 32$
- $32 \div 2 = 16$
- $16 \div 2 = 8$
- $8 \div 2 = 4$
- $4 \div 2 = 2$
- $2 \div 2 = 1$
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:
$(2^6)^{\frac{1}{2}}$
Step 3: Applying the Power of a Power Property
[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:
$2 \times 2 \times 2 = 8$
Step 4: Method 2 - Perfect Square Identification (Direct Approach)
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 $(a^m)^n = a^{m \cdot n}$:
$8^{2 \times \frac{1}{2}} = 8^1 = 8$
Geometric Representation of the Square Root
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 $64$.
Final Conclusion
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
More Questions from Class 9 Mathematics Number Systems EXERCISE 1.5
- Q1(ii): Find : (ii) $32^{\frac{1}{5}}$
- Q1(iii): Find : (iii) $125^{\frac{1}{3}}$
- Q2(i): Find : (i) $9^{\frac{3}{2}}$
- Q2(ii): Find : (ii) $32^{\frac{2}{5}}$
- Q2(iii): Find : (iii) $16^{\frac{3}{4}}$
- Q2(iv): Find : (iv) $125^{-\frac{1}{3}}$
- Q3(i): Simplify : (i) $2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}$
- Q3(ii): Simplify : (ii) $(\frac{1}{3^3})^7$
- Q3(iii): Simplify : (iii) $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$
- Q3(iv): Simplify : (iv) $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}$
CBSE Solutions for Class 9 Mathematics Number Systems
Chapters in CBSE - Class 9 Mathematics
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