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## Note on Indices

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An indices is a number with the power. For example: am; a is called the base and m is the power.

In 4x3, its coefficient is 4, base is x and power is 3. The power of the base of an algebraic term is also called index. The plural form of index is indices.

### Law of indices

1. $$a^m \times a^n = a^{m+n}$$
2. $$a^m ÷ a^n = a^{m - n}$$
3. $$(a^m)^n = a^{mn}$$
4. $$\left ( \frac {a} {b} \right ) ^m = \frac {a^m} {b^m}$$
5. $$\sqrt [n] {a^m} = a^ {\frac {m} {n} }$$
6. $$a^0 = 1$$
7. $$\left( \frac {a} {b} \right)^{-m} =\left( \frac {b} {a} \right)^m$$
8. $$If \: a^m = a^n \: then, \: m = n$$
9. $$a^{-m} = \frac {1}{a^m} \: (a ≠ o )$$
10. $$\left( \sqrt [n] {a} \right )^n = \left ( a^{\frac {1}{n}} \right ) ^n = a$$
11. $$\sqrt [n]{a} . \sqrt [n]{b} = \sqrt [n]{ab}$$
12. $$\sqrt [m]{\sqrt [n] {a}} = \sqrt [mn]{a}$$
13. $$\frac {\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\frac{a}{b}} = \left( \frac{a}{b} \right ) ^{\frac{1}{n}}$$
14. $$1^m =1$$, when 'm' is a whole number.
15. $$a^x = b⇒ a = b^{\frac{1}{x}}$$ [x ≠ 0]

Note:It is incorrect to write $$\sqrt{16} = \pm 4$$ because $$\sqrt{16}$$ denotes the principle or positive square root of 16.

1. $$a^m \times a^n = a^{m+n}$$
2. $$a^m ÷ a^n = a^{m - n}$$
3. $$(a^m)^n = a^{mn}$$
4. $$\left ( \frac {a} {b} \right ) ^m = \frac {a^m} {b^m}$$
5. $$\sqrt [n] {a^m} = a^ {\frac {m} {n} }$$
6. $$a^0 = 1$$
.

### Very Short Questions

Solution:

\begin{align*} &= \frac{14^6\times15^5}{35^6\times6^5}\\&=\frac{2^6\times7^6\times3^5\times5^5}{5^6\times7^6\times2^5\times3^5}=\frac{2^{6-5}}{5^{6-5}}=\frac{2}{5}\:_\text {Ans.} \end{align*}

Solution:

\begin{align*} &= \sqrt [3] {\sqrt{(64)^{-1}}}\\&=\sqrt[3] {\sqrt{\frac{1} {64}}}=\sqrt [3] {(\frac{1}{8})^{2\times \frac{1}{2}}}=(\frac{1}{2})^{3\times \frac{1}{3}}=\frac{1}{2}\:_\text {Ans.} \end{align*}

Solution:

\begin{align*} &= \left( \frac {x^0} {64} \right)^{-\frac{2}{3}}=\frac{1}{\left( \frac{x^0}{64}\right)^{\frac{2}{3}}} =\frac{1}{\left(\frac{1}{4}\right)^{3\times \frac{2}{3}}}=\frac{1}{(\frac{1}{4})^2} =\frac{1}{\frac{1}{16}}=16\:_\text {Ans.} \end{align*}

Solution:

\begin{align*} &=\sqrt[6]{\left (\frac{1}{64}\right)^{-1}}= \sqrt[6]{64}= 64^{\frac{1}{6}}=2^{6\times \frac{1}{6}}=2 \:_\text {Ans.} \end{align*}

Solution:

\begin{align*} &= \left (\frac{8}{27} \right)^{-\frac{2}{3}}=\frac{1}{\left (\frac{8}{27}\right)^{\frac{2}{3}}}= \frac{1}{ \left (\frac{2}{3} \right)^{3 \times \frac {2}{3}}}= \frac{1}{ \left (\frac{2}{3} \right)^2}=\frac{1}{\frac{4}{9}}= \frac{9}{4} :_\text {Ans.} \end{align*}

Solution:

\begin{align*} &= \left (4^\frac{3}{4} \right)^{-\frac{4}{3}}=\frac{1}{{4^{\frac{3}{4}}}\times\frac{4}{3}}=\frac{1}{4} :_\text {Ans.} \end{align*}

Solution:

\begin{align*} &= \left (\frac{8}{27} \right)^{-\frac{2}{3}}=\frac{1}{\left (\frac{8}{27}\right)^{\frac{2}{3}}}= \frac{1}{ \left (\frac{2}{3} \right)^{3 \times \frac {2}{3}}}= \frac{1}{ \left (\frac{2}{3} \right)^2}=\frac{1}{\frac{4}{9}}= \frac{9}{4} :_\text {Ans.} \end{align*}

Solution:

\begin{align*} &= \left (\frac{1}{27} \right)^{-\frac{4}{3}}\times \left (27\right) ^{\frac{1}{3}}= \left(\frac{1}{3^3} \right)^{-\frac{4}{3}}\times3^{3\times \frac{1}{3}} = 3^{-3 \times \frac{-4}{3}}\times3= 3^4 \times 3 = 81\times3=243:_\text {Ans.} \end{align*}

Solution:

\begin{align*} &=\left (\frac{8}{27} \right)^{-\frac{2}{3}}=3^{4\frac{3}{4}} \times6^{3 \times \frac{-2}{3}}=3^3 \times6^{-2}=\frac{27}{6^2}=\frac{27}{36}=\frac{3}{4}:_\text {Ans.} \end{align*}

Solution:

\begin{align*} &= \left(\frac{27}{8} \right)^{-\frac{1}{3}} {\left[\left(\frac{81}{16}\right)^{\frac{1}{4}} \div \left(\frac{4}{25}\right )^{-\frac{1}{2}}\right]}\\&= \left(\frac{3}{2} \right)^ {3 \times\frac{-1}{3}} \left[\left(\frac{3}{4}\right )^{4 \times \frac{1}{4}} \div \left(\frac{2}{5}\right)^{2\times\frac{-1}{2}}\right ]\\&=\left(\frac{3}{2}^{-1}\right)\left[\frac{3}{4} \div \frac{1}{\frac{2}{5}}\right]\\&=\frac{1}{\left(\frac{3}{2}\right)}\left[\frac{3}{4}\times\frac{2}{5} \right ]\\&=\frac{2}{3} \times \frac{3}{2} \times\frac{1}{5}\\&=\frac{1}{5} _\text {Ans.} \end{align*}

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