Indices

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.

 

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  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. \( a^0 = 1 \)

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*}

Solution:

Here, \(\sqrt[\frac 1{bc}]{\frac{x^{\frac 1b}}{x^{\frac 1c}}}\) × \(\sqrt[\frac 1{ab}]{\frac{x^{\frac 1a}}{x^{\frac 1b}}}\) × \(\sqrt[\frac 1{ca}]{\frac{x^{\frac 1c}}{x^{\frac 1a}}}\)

 = \(\frac{x^{\frac{1}{b}÷{\frac{1}{bc}}}}{x^{\frac{1}{c}÷{\frac{1}{bc}}}}\) × \(\frac{x^{\frac{1}{a}÷{\frac{1}{ab}}}}{x^{\frac{1}{b}÷{\frac{1}{ab}}}}\) × \(\frac{x^{\frac{1}{c}÷{\frac{1}{ca}}}}{x^{\frac{1}{a}÷{\frac{1}{ca}}}}\)

 = \(\frac{x^{\frac{bc}{b}}}{x^{\frac{bc}{c}}}\) × \(\frac{x^{\frac{ab}{a}}}{x^{\frac{ab}{b}}}\) × \(\frac{x^{\frac{ca}{c}}}{x^{\frac{ca}{a}}}\)

 = \(\frac{x^{c}}{x^{b}}\) × \(\frac{x^{b}}{x^{a}}\) × \(\frac{x^{a}}{x^{c}}\) = 1 ans.

Solution:

Here,  \(\sqrt[bc]{\frac{x^{\frac bc}}{x^{\frac cb}}}\) ×  \(\sqrt[ca]{\frac{x^{\frac ca}}{x^{\frac ac}}}\) ×  \(\sqrt[ab]{\frac{x^{\frac ab}}{x^{\frac ba}}}\)

 = \(\frac{x^{\frac{b}{c}×{\frac{1}{bc}}}}{x^{\frac{c}{a}×{\frac{1}{ca}}}}\) × \(\frac{x^{\frac{a}{b}×{\frac{1}{ab}}}}{x^{\frac{b}{a}×{\frac{1}{ab}}}}\) × \(\frac{x^{\frac{c}{a}×{\frac{1}{ca}}}}{x^{\frac{a}{c}×{\frac{1}{ca}}}}\)

 = \(\frac{x^{\frac{1}{c^2}}}{x^{\frac{1}{b^2}}}\) × \(\frac{x^{\frac{1}{a^2}}}{x^{\frac{1}{c^2}}}\) × \(\frac{x^{\frac{1}{b^2}}}{x^{\frac{1}{a^2}}}\)

 = x\(\frac{1}{c^2}\) - \(\frac{1}{b^2}\) × x\(\frac{1}{a^2}\) - \(\frac{1}{c^2}\) × x\(\frac{1}{b^2}\) - \(\frac{1}{a^2}\)

 = x\(\frac{1}{c^2}\) - \(\frac{1}{b^2}\) + \(\frac{1}{a^2}\) - \(\frac{1}{c^2}\) + \(\frac{1}{b^2}\) - \(\frac{1}{a^2}\)

 = x0 = 1 ans.

Solution: 

Here, \(\frac{1}{1+a^{x-y}+a^{z-y}}\) + \(\frac{1}{1+a^{y-z}+a^{x-z}}\) + \(\frac{1}{1+a^{z-x}+a^{y-x}}\)

 = \(\frac{1}{1+{\frac{a^x}{a^y}}+{\frac{a^z}{a^y}}}\) + \(\frac{1}{1+{\frac{a^y}{a^z}}+{\frac{a^x}{a^z}}}\) + \(\frac{1}{1+{\frac{a^z}{a^x}}+{\frac{a^y}{a^x}}}\)

 = \(\frac{1}{\frac{a^y+a^x+a^z}{a^y}}\) +  \(\frac{1}{\frac{a^z+a^y+a^x}{a^z}}\) +  \(\frac{1}{\frac{a^x+a^z+a^y}{a^x}}\)

 = \(\frac{a^y}{a^x+a^y+a^z}\) +  \(\frac{a^z}{a^x+a^y+a^z}\) +  \(\frac{a^x}{a^x+a^y+a^z}\)

 =  \(\frac{a^x+a^y+a^z}{a^x+a^y+a^z}\) = 1 ans.

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  • (left(frac{25}{16} ight)^{-1/2}) ([left(frac{125}{64} ight)^{1/3}  ÷ left(frac{125}{64} ight)^{-1/3}])

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  • (frac{5^{m+2}-5^m}{5^{m+1} + 5^m})

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  • (frac{6^{n+2}-6^n}{6^{n+1} + 6^n})

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  • (frac{2^{x+4}-2^x}{5.2^x})

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  • (frac{3^{x+2}+3^x}{5.3^x})

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  • (frac{5^{n+2}-2.5^n}{23.5^n})

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  • (frac {4^{m+4}+ 4^{m+1}}{4^{m+2}-4^m})

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  • (frac{5^x-2.5^{x-1}}{3.5^x})

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  • (frac{2^x-2^{x-2}}{3.2^x})

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  • (frac{3^{3a+2}-3^{3a+1}}{6×27^a})

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  • (ax÷ay)x2+xy+y2.(ay÷az)y2+yz=z2.(az÷ax)z2=zx+x2

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  • (64x3÷27a-3)-2/3

    9/6a2x2


    9/16a2x2


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    9/16ax2


  • (64x3÷27a-3)-2/3

    9/16a2x2


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    8/16a2x2


    9/16ax2


  • (8x3÷27a-3)-2/3

    8/5a2x2
    9/4a2x2
    9/3ax2
    9/5a2x2
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samjhana thapa

3^x×9 9^(-x)

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5^(x-2) × 3^(2x-3) =135

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Ask any queries on this note.x^3±y^3=?

Ask any queries on this note.x^3_Y3=?