Examples on Indices and Surds

Examples on Indices and Surds

Indices and Surds:Theory, Concepts and formula

Example: 1 For $x\ne 0,{{\left( \frac{{{x}^{l}}}{{{x}^{m}}} \right)}^{({{l}^{2}}+lm+{{m}^{2}})}}$${{\left( \frac{{{x}^{m}}}{{{x}^{n}}} \right)}^{({{m}^{2}}+nm+{{n}^{2}})}}{{\left( \frac{{{x}^{n}}}{{{x}^{l}}} \right)}^{({{n}^{2}}+nl+{{l}^{2}})}}$=


Solution: ${{\left( \frac{{{x}^{l}}}{{{x}^{m}}} \right)}^{{{l}^{2}}+lm+{{m}^{2}}}}{{\left( \frac{{{x}^{m}}}{{{x}^{n}}} \right)}^{{{m}^{2}}+nm+{{n}^{2}}}}{{\left( \frac{{{x}^{n}}}{{{x}^{l}}} \right)}^{{{n}^{2}}+nl+{{l}^{2}}}}$


=${{({{x}^{l-m}})}^{({{l}^{2}}+lm+{{m}^{2}})}}{{({{x}^{m-n}})}^{{{m}^{2}}+nm+{{n}^{2}}}}$${{({{x}^{n-l}})}^{{{n}^{2}}+nl+{{l}^{2}}}}$=${{x}^{{{l}^{3}}-{{m}^{3}}}}.{{x}^{{{m}^{3}}-{{n}^{3}}}}.{{x}^{{{n}^{3}}-{{l}^{3}}}}$=${{x}^{{{l}^{3}}-{{m}^{3}}+{{m}^{3}}-{{n}^{3}}+{{n}^{3}}-{{l}^{3}}}}={{x}^{0}}$=1


Example: 2 If ${{2}^{x}}={{4}^{y}}={{8}^{z}}$and $xyz=288,$then$\frac{1}{2x}+\frac{1}{4y}+\frac{1}{8z}=$

Solution: ${{2}^{x}}={{2}^{2y}}={{2}^{3z}}\,i.e.,\,x=2y=3z=k$ (say). Then $xyz=\frac{{{k}^{3}}}{6}=288$, So $k=12$

$\therefore x=12,y=6,z=4$. Therefore, $\frac{1}{2x}+\frac{1}{4y}+\frac{1}{8z}=\frac{11}{96}$


Example: 3 Simplify $\frac{{{2.3}^{n+1}}+{{7.3}^{n-1}}}{{{3}^{n+2}}-2{{(1/3)}^{l-n}}}=$

Solution: $\frac{{{2.3}^{n+1}}+{{7.3}^{n-1}}}{{{3}^{n+2}}-2{{\left( \frac{1}{3} \right)}^{1-n}}}=\frac{{{2.3}^{n-1}}{{.3}^{2}}+{{7.3}^{n-1}}}{{{3}^{n-1}}{{.3}^{3}}-{{2.3}^{n-1}}}$$=\frac{{{3}^{n-1}}[18+7]}{{{3}^{n-1}}[27-2]}=1$


Example: 4 If ${{\left( \frac{2}{3} \right)}^{x+2}}={{\left( \frac{3}{2} \right)}^{2-2x}},$then x =

Solution: ${{\left( \frac{2}{3} \right)}^{x+2}}={{\left( \frac{3}{2} \right)}^{2-2x}}$ or ${{\left( \frac{2}{3} \right)}^{x+2}}={{\left( \frac{2}{2} \right)}^{2-2x}}$. Clearly $x+2=2x-2$ Þ $x=4$


Example: 5 Solve the equation ${{4}^{({{x}^{2}}+2)}}-{{9.2}^{({{x}^{2}}+2)}}+8=0$

Solution: ${{4}^{({{x}^{2}}+2)}}-{{9.2}^{({{x}^{2}}+2)}}+8=0$ $\Rightarrow {{\left( {{2}^{({{x}^{2}}+2)}} \right)}^{2}}-{{9.2}^{({{x}^{2}}+2)}}+8=0$

Put ${{2}^{({{x}^{2}}+2)}}^{2}=y$. Then ${{y}^{2}}-9y+8=0$, which gives $y=8,y=1$

When $y=8\,\,\Rightarrow \,\,{{2}^{{{x}^{2}}+2}}=8$ or ${{2}^{{{x}^{2}}+2}}={{2}^{3}}$ or ${{x}^{2}}+2=3$ or ${{x}^{2}}=1$ or $x=1,-1$

When $y=1$ or ${{2}^{{{x}^{2}}+2}}=1$ or ${{2}^{{{x}^{2}}+2}}={{2}^{o}}$ or ${{x}^{2}}+2=0$ or ${{x}^{2}}=-2$, which is not possible.


Example: 6 The greatest number among $\sqrt[3]{9},\sqrt[4]{11},\sqrt[6]{17}$is

Solution: $\sqrt[3]{9},\,\,\sqrt[4]{11},\,\,\sqrt[6]{17}$

$\because L.C.M$of 3, 4, 6 is 12

$\therefore \sqrt[3]{9}={{9}^{1/3}}={{({{9}^{4}})}^{1/12}}={{(6561)}^{1/12}}$, $\sqrt[4]{11}={{(11)}^{1/4}}{{({{11}^{3}})}^{1/12}}={{(1331)}^{1/12}}$, $\sqrt[6]{17}={{(17)}^{1/6}}={{({{17}^{2}})}^{1/2}}={{(289)}^{1/12}}$

Hence $\sqrt[3]{9}$is the greatest number.


Example: 7 Solve the equation $\sqrt{(x+1)}-\sqrt{(x-1)}=\sqrt{(4x-1)}$, $x\in R$

Solution: Given $\sqrt{(x+1)}-\sqrt{(x-1)}=\sqrt{(4x-1)}$ .....(i)

Squaring both sides, we get, $-2\sqrt{({{x}^{2}}-1)}=2x-1$

Squaring again, we get, $x=\frac{5}{4},$ which does not satisfy equation (i)

Hence, there is no solution of the given equation.