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Q2(iv):
Find :
(iv) $125^{-\frac{1}{3}}$
Solution :
Given Expression & Initial Setup
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.
Step 1: Prime Factorization of the Base
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 ($5$):
- $125 \div 5 = 25$
- $25 \div 5 = 5$
- $5 \div 5 = 1$
Thus, the prime factorization is:
$125 = 5 \times 5 \times 5 = 5^3$
Step 2: Substitution into the Original Expression
We substitute the prime factored form of the base back into the original expression.
$125^{-\frac{1}{3}} = (5^3)^{-\frac{1}{3}}$
Step 3: Application of the Power of a Power Rule
[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 ($-\frac{1}{3}$):
$(5^3)^{-\frac{1}{3}} = 5^{3 \times \left(-\frac{1}{3}\right)}$
Step 4: Simplifying the Exponent
Perform the multiplication in the exponent:
$3 \times \left(-\frac{1}{3}\right) = -\frac{3}{3} = -1$
Substituting this back yields:
$5^{-1}$
Step 5: Application of the Negative Exponent Rule
[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 $\frac{1}{5}$.
More Questions from Class 9 Mathematics Number Systems EXERCISE 1.5
- Q1(i): Find : (i) $64^{\frac{1}{2}}$
- 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) $16^{\frac{3}{4}}$
- 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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