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# 2.Simplify and express the result in power notation with a positive exponent. (i)$$(-4)^5÷(-4)^8$$ (ii)$$(\frac{1}{2^3})^2$$ (iii)$$(-3)^4\times (\frac{5}3)^4$$ (iv)$$(3^{-7}÷3^{-10})\times 3^{-5}$$ (v)$$2^{-3}\times(-7)^{-3}$$

(i)We have:$$(-4)^5÷(-4)^8$$
$$=(-4)^{5-8}=(-4)^{-3}=\frac{1}{(-4)^3}\quad$$[∵$$a^{m}÷a^{n}=a^{m-n}$$]

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

(ii)We have:$$(\frac{1}{2^3})^2$$

$$(\frac{1}{2^3})^2=\frac{(1)^2}{(2^3)^2}=\frac{1}{2^6}=(\frac{1}2)^6$$

(iii)We have:$$(-3)^4\times (\frac{5}3)^4$$

$$=(-3)^4\times (\frac{5}3)^4=(-3)^4\times \frac{(5)^4}{(3)^4}$$

$$=\frac{(3)^4\times(5)^4}{(3)^4}=(5)^4$$

(iv)We have:$$(3^{-7}÷3^{-10})\times 3^{-5}$$

$$=(3^{-7}÷3^{-10})\times 3^{-5}=3^{-7-(-10)}\times 3^{-5}$$

$$=3^{-7+10}\times 3^{-5}=3^3\times 3^{-5}=3^{(3-5)}$$

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

(v)We have:$$2^{-3}\times(-7)^{-3}$$

$$=2^{-3}\times(-7)^{-3}=[2\times(-7)]^{-3}=(-14)^{-3}$$

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