CBSE Class 8 Maths Chapter 12: Exponents and Powers NCERT Solutions

We need to find the value of the given expressions involving negative exponents. We will use the law of exponents $a^{-m} = \frac{1}{a^m}$.

1. For $3^\{-2\}$:

   $$3^\{-2\}\; =\; \backslash frac\{1\}\{3^2\}\; =\; \backslash frac\{1\}\{9\}$$

   Thus, the value of $3^\{-2\}$ is $\backslash frac\{1\}\{9\}$.

2. For $(-4)^\{-2\}$:

   $$(-4)^\{-2\}\; =\; \backslash frac\{1\}\{(-4)^2\}$$

   Since $(-4)^2 = (-4) \times (-4) = 16$, we have:

   $$=\; \backslash frac\{1\}\{16\}$$

   Thus, the value of $(-4)^\{-2\}$ is $\backslash frac\{1\}\{16\}$.

3. For $\backslash left(\backslash frac\{1\}\{2\}\backslash right)^\{-5\}$:

   Using the rule $\left(\frac{a}{b}\right)^{-m} = \left(\frac{b}{a}\right)^{m}$, we get:

   $$\backslash left(\backslash frac\{1\}\{2\}\backslash right)^\{-5\}\; =\; \backslash left(\backslash frac\{2\}\{1\}\backslash right)^\{5\}$$

   Now, we calculate $2^5$:

   $$=\; 2^5\; =\; 2\; \backslash times\; 2\; \backslash times\; 2\; \backslash times\; 2\; \backslash times\; 2\; =\; 32$$

   Thus, the value of $\backslash left(\backslash frac\{1\}\{2\}\backslash right)^\{-5\}$ is 32.

We will simplify each expression using the laws of exponents.

1. For $(-4)^5\; \backslash div\; (-4)^8$:

   Using the quotient law $a^m \div a^n = a^{m-n}$:

   $(-4)^5\; \backslash div\; (-4)^8\; =\; (-4)^\{5-8\}\; =\; (-4)^\{-3\}$

   To express this with a positive exponent, we use $a^{-m} = \frac{1}{a^m}$:

   $=\; \backslash frac\{1\}\{(-4)^3\}$

   The simplified form with a positive exponent is $\backslash frac\{1\}\{(-4)^3\}$.

2. For $\backslash left(\backslash frac\{1\}\{2^3\}\backslash right)^2$:

   Using the power of a power law $(a^m)^n = a^{m \times n}$ and the rule $\left(\frac{a}{b}\right)^m = \frac{a^m}{b^m}$:

   $\backslash left(\backslash frac\{1\}\{2^3\}\backslash right)^2\; =\; \backslash frac\{1^2\}\{(2^3)^2\}\; =\; \backslash frac\{1\}\{2^\{3\; \backslash times\; 2\}\}\; =\; \backslash frac\{1\}\{2^6\}$

   The simplified form with a positive exponent is $\backslash frac\{1\}\{2^6\}$.

3. For $\backslash left(-3\backslash right)^4\; \backslash times\; \backslash left(\backslash frac\{5\}\{3\}\backslash right)^4$:

   Using the rule $(ab)^m = a^m b^m$ and $\left(\frac{a}{b}\right)^m = \frac{a^m}{b^m}$:

   $\backslash left(-3\backslash right)^4\; \backslash times\; \backslash left(\backslash frac\{5\}\{3\}\backslash right)^4\; =\; \backslash left(-3\backslash right)^4\; \backslash times\; \backslash frac\{5^4\}\{3^4\}$

   Since $(-3)^4 = 3^4$ (because any negative number raised to an even power becomes positive):

   $=\; 3^4\; \backslash times\; \backslash frac\{5^4\}\{3^4\}$

   Now, we can cancel out $3^4$:

   $=\; 5^4$

   The simplified form with a positive exponent is $5^4$.

4. For $(3^\{-7\}\; \backslash div\; 3^\{-10\})\; \backslash times\; 3^\{-5\}$:

   First, apply the quotient law $a^m \div a^n = a^{m-n}$:

   $(3^\{-7\}\; \backslash div\; 3^\{-10\})\; =\; 3^\{-7\; -\; (-10)\}\; =\; 3^\{-7\; +\; 10\}\; =\; 3^3$

   Now, multiply this result by $3^{-5}$ using the product law $a^m \times a^n = a^{m+n}$:

   $3^3\; \backslash times\; 3^\{-5\}\; =\; 3^\{3\; +\; (-5)\}\; =\; 3^\{3-5\}\; =\; 3^\{-2\}$

   To express this with a positive exponent, use $a^{-m} = \frac{1}{a^m}$:

   $=\; \backslash frac\{1\}\{3^2\}$

   The simplified form with a positive exponent is $\backslash frac\{1\}\{3^2\}$.

5. For $2^\{-3\}\; \backslash times\; (-7)^\{-3\}$:

   Using the rule $(ab)^m = a^m b^m$:

   $2^\{-3\}\; \backslash times\; (-7)^\{-3\}\; =\; \backslash \{2\; \backslash times\; (-7)\backslash \}^\{-3\}\; =\; (-14)^\{-3\}$

   To express this with a positive exponent, use $a^{-m} = \frac{1}{a^m}$:

   $=\; \backslash frac\{1\}\{(-14)^3\}$

   The simplified form with a positive exponent is $\backslash frac\{1\}\{(-14)^3\}$.

We will evaluate each expression using the laws of exponents.

1. For $\backslash left(\backslash frac\{1\}\{2\}\backslash right)^\{-2\}\; +\; \backslash left(\backslash frac\{1\}\{3\}\backslash right)^\{-2\}\; +\; \backslash left(\backslash frac\{1\}\{4\}\backslash right)^\{-2\}$:

   Using the rule $\left(\frac{a}{b}\right)^{-m} = \left(\frac{b}{a}\right)^{m}$:

   $$\backslash left(\backslash frac\{1\}\{2\}\backslash right)^\{-2\}\; =\; \backslash left(\backslash frac\{2\}\{1\}\backslash right)^\{2\}\; =\; 2^2\; =\; 4$$

   $$\backslash left(\backslash frac\{1\}\{3\}\backslash right)^\{-2\}\; =\; \backslash left(\backslash frac\{3\}\{1\}\backslash right)^\{2\}\; =\; 3^2\; =\; 9$$

   $$\backslash left(\backslash frac\{1\}\{4\}\backslash right)^\{-2\}\; =\; \backslash left(\backslash frac\{4\}\{1\}\backslash right)^\{2\}\; =\; 4^2\; =\; 16$$

   Adding these values:

   $$4\; +\; 9\; +\; 16\; =\; 29$$

   The value of the expression is 29.

2. For $(3^0\; +\; 4^\{-1\})\; \backslash times\; 2^2$:

   We know that $a^0 = 1$ and $a^{-m} = \frac{1}{a^m}$:

   $$(3^0\; +\; 4^\{-1\})\; \backslash times\; 2^2\; =\; \backslash left(1\; +\; \backslash frac\{1\}\{4\}\backslash right)\; \backslash times\; 2^2$$

   First, calculate the sum inside the parenthesis:

   $$\backslash left(1\; +\; \backslash frac\{1\}\{4\}\backslash right)\; =\; \backslash left(\backslash frac\{4\}\{4\}\; +\; \backslash frac\{1\}\{4\}\backslash right)\; =\; \backslash frac\{5\}\{4\}$$

   Now, multiply by $2^2$:

   $$=\; \backslash frac\{5\}\{4\}\; \backslash times\; 2^2\; =\; \backslash frac\{5\}\{4\}\; \backslash times\; 4\; =\; 5$$

   The value of the expression is 5.

3. For $(2^\{-1\}\; \backslash times\; 4^\{-1\})\; \backslash div\; 2^\{-2\}$:

   Using $a^{-m} = \frac{1}{a^m}$:

   $$(2^\{-1\}\; \backslash times\; 4^\{-1\})\; =\; \backslash left(\backslash frac\{1\}\{2\}\; \backslash times\; \backslash frac\{1\}\{4\}\backslash right)\; =\; \backslash frac\{1\}\{8\}$$

   We can write $\frac{1}{8}$ as $2^{-3}$ since $8 = 2^3$. So, the expression becomes:

   $$2^\{-3\}\; \backslash div\; 2^\{-2\}$$

   Using the quotient law $a^m \div a^n = a^{m-n}$:

   $$=\; 2^\{-3\; -\; (-2)\}\; =\; 2^\{-3\; +\; 2\}\; =\; 2^\{-1\}$$

   Using $a^{-m} = \frac{1}{a^m}$ again:

   $=\; \backslash frac\{1\}\{2\}$

   The value of the expression is $\backslash frac\{1\}\{2\}$.

4. For $\backslash left(3^\{-1\}\; +\; 4^\{-1\}\; +\; 5^\{-1\}\backslash right)^0$:

   Any non-zero number raised to the power of 0 is 1. The expression inside the parenthesis is:

   $\backslash left(\backslash frac\{1\}\{3\}\; +\; \backslash frac\{1\}\{4\}\; +\; \backslash frac\{1\}\{5\}\backslash right)$

   This sum is a non-zero number. Therefore, raising it to the power of 0 gives:

   $$\backslash left(\backslash frac\{1\}\{3\}\; +\; \backslash frac\{1\}\{4\}\; +\; \backslash frac\{1\}\{5\}\backslash right)^0\; =\; 1$$

   The value of the expression is 1.

5. For $\backslash left\backslash \{\backslash left(\backslash frac\{-2\}\{3\}\backslash right)^\{-2\}\backslash right\backslash \}^2$:

   First, evaluate the inner part $\backslash left(\backslash frac\{-2\}\{3\}\backslash right)^\{-2\}$ using the rule $\left(\frac{a}{b}\right)^{-m} = \left(\frac{b}{a}\right)^{m}$:

   $$\backslash left(\backslash frac\{-2\}\{3\}\backslash right)^\{-2\}\; =\; \backslash left(\backslash frac\{3\}\{-2\}\backslash right)^\{2\}\; =\; \backslash left(\backslash frac\{3\}\{2\}\backslash right)^\{2\}$$

   Now, raise this result to the power of 2:

   $$\backslash left\backslash \{\backslash left(\backslash frac\{3\}\{2\}\backslash right)^\{2\}\backslash right\backslash \}^2$$

   Using the power of a power law $(a^m)^n = a^{m \times n}$:

   $$=\; \backslash left(\backslash frac\{3\}\{2\}\backslash right)^\{2\; \backslash times\; 2\}\; =\; \backslash left(\backslash frac\{3\}\{2\}\backslash right)^\{4\}$$

   Calculating the value:

   $$=\; \backslash frac\{3^4\}\{2^4\}\; =\; \backslash frac\{81\}\{16\}$$

   The value of the expression is $\backslash frac\{81\}\{16\}$.