Surds and Indices - 01

Explanation:-

$$\eqalign{ & {\left( {32 \times {{10}^{ - 5}}} \right)^{ - 2}} \times 64 \div \left( {{2^{16}} \times {{10}^{ - 4}}} \right) = {10^?} \cr & \Rightarrow {\left( {{2^5} \times {{10}^{ - 5}}} \right)^{ - 2}} \times {2^6} \div \left( {{2^{16}} \times {{10}^{ - 4}}} \right) \cr & \,\,\,\,\,\,\,\,\, = {10^?}.....\left[ {\because {{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right] \cr & \Rightarrow \frac{{{2^{10}} \times {{10}^{ - 10}} \times {2^6}}}{{{2^{16}} \times {{10}^{ - 4}}}} \cr & \,\,\,\,\,\,\,\,\,\, = {10^?}.....\left[ {\because {a^m} \times {a^n} = {a^{m + n}}} \right] \cr & \Rightarrow \frac{{{2^{16}} \times {{10}^4}}}{{{2^{16}} \times {{10}^{10}}}} \cr & \,\,\,\,\,\,\,\, = {10^?}.....\left[ {{a^{ - m}} = \frac{1}{{{a^m}}}} \right] \cr & \Rightarrow {10^{4 - 10}} = {10^?} \cr & \Rightarrow {10^{ - 6}} = {10^?} \cr & \Rightarrow ? = - 6 \cr}$$

Explanation:-

$$\eqalign{ & {21^?} \times {21^{6.5}} = {21^{12.4}}.....\left[ {\because {a^m} \times {a^n} = {a^{m + n}}} \right] \cr & \Rightarrow {21^{? + 6.5}} = {21^{12.4}} \cr & \Rightarrow ? + 6.5 = 12.4 \cr & \Rightarrow ? = 12.4 - 6.5 \cr & \Rightarrow ? = 5.9 \cr}$$

Explanation:-

$$\eqalign{ & \frac{1}{{1 + \sqrt 2 + \sqrt 3 }} + \frac{1}{{1 - \sqrt 2 + \sqrt 3 }} \cr & = \frac{1}{{1 + \sqrt 3 + \sqrt 2 }} + \frac{1}{{1 + \sqrt 3 - \sqrt 2 }} \cr & = \frac{{1 + \sqrt 3 - \sqrt 2 + 1 + \sqrt 3 + \sqrt 2 }}{{{{\left( {1 + \sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr & = \frac{{2 + 2\sqrt 3 }}{{4 + 2\sqrt 3 - 2}} \cr & = \frac{{2 + 2\sqrt 3 }}{{2 + 2\sqrt 3 }} \cr & = 1 \cr}$$

Explanation:-

$$\eqalign{ & {\text{The exponential form of }} \cr & \sqrt {\sqrt 2 \times \sqrt 3 } \cr & = \sqrt {{6^{\frac{1}{2}}}} \cr & = {\left( {{6^{\frac{1}{2}}}} \right)^{\frac{1}{2}}} \cr & = {6^{\frac{1}{4}}} \cr}$$

Explanation:-

$$\eqalign{ & \frac{{\left( {3x - 2y} \right)}}{{\left( {3x + 2y} \right)}} = \frac{5}{6} \cr & \Rightarrow 18x - 12y = 10x + 15y \cr & \Rightarrow 8x = 27y \cr & \Rightarrow \frac{x}{y} = \frac{{27}}{8} \cr & \Rightarrow {\left( {\frac{{\root 3 \of x + \root 3 \of y }}{{\root 3 \of x - \root 3 \of y }}} \right)^2} \cr & \Rightarrow {\left( {\frac{{\root 3 \of {27} + \root 3 \of 8 }}{{\root 3 \of {27} - \root 3 \of 8 }}} \right)^2} \cr & \Rightarrow {\left( {\frac{{3 + 2}}{{3 - 2}}} \right)^2} \cr & \Rightarrow {\left( 5 \right)^2} \cr & \Rightarrow 25 \cr}$$

Explanation:-

$$\eqalign{ & {{\text{2}}^{0.7x}}{\text{.}}{{\text{3}}^{ - 1.25y}}{\text{ = }}\frac{{8\sqrt 6 }}{{27}} \cr & \Leftrightarrow \frac{{{{\text{2}}^{0.7x}}}}{{{{\text{3}}^{ - 1.25y}}}}{\text{ = }}\frac{{{2^3}{{.2}^{\frac{1}{2}}}{{.2}^{\frac{1}{2}}}}}{{{3^3}}} \cr & \Leftrightarrow \frac{{{2^{\left( {3 + \frac{1}{2}} \right)}}}}{{{3^{\left( {3 + \frac{1}{2}} \right)}}}} = \frac{{{2^{\frac{7}{2}}}}}{{{2^{\frac{5}{2}}}}} = \frac{{{2^{3.5}}}}{{{3^{2.5}}}} \cr & \therefore 0.7x = 3.5 \Rightarrow x = \frac{{3.5}}{{0.7}}{\text{ = 5}} \cr & {\text{and }}1.25y = 2.5 \cr & \Rightarrow y = \frac{{2.5}}{{1.25}} = 2 \cr}$$

Explanation:-

$$\eqalign{ & {{\text{2}}^x}{\text{ = }}{{\text{8}}^{y + 1}} \cr & \Leftrightarrow {{\text{2}}^x}{\text{ = }}{\left( {{2^3}} \right)^{y + 1}} = {2^{\left( {3y + 3} \right)}} \cr & \Leftrightarrow x = 3y + 3 \cr & \Leftrightarrow x - 3y = 3.......(i) \cr & {9^y} = {3^{x - 9}} \cr & \Leftrightarrow {\left( {{3^2}} \right)^y}{\text{ = }}{{\text{3}}^{x - 9}} \cr & \Leftrightarrow 2y = x - 9 \cr & \Leftrightarrow x - 2y = 9......({\text{ii}}) \cr & Subtracting(i)from(ii), \cr & we\,get\,y = 6 \cr & {\text{Putting }}y\,{\text{ = 6 in (i),}} \cr & {\text{we get }}x{\text{ = 21}} \cr & \therefore x + y = 21 + 6 = 27 \cr}$$

Explanation:-

$$\eqalign{ & \frac{{\sqrt {10 + \sqrt {25 + \sqrt {108 + \sqrt {154 + \sqrt {225} } } } } }}{{\root 3 \of 8 }} \cr & \Rightarrow \frac{{\sqrt {10 + \sqrt {25 + \sqrt {108 + \sqrt {169} } } } }}{2} \cr & \Rightarrow \frac{{\sqrt {10 + \sqrt {25 + \sqrt {121} } } }}{2} \cr & \Rightarrow \frac{{\sqrt {10 + \sqrt {36} } }}{2} \cr & \Rightarrow \frac{{\sqrt {16} }}{2} \cr & \Rightarrow \frac{4}{2} \cr & \Rightarrow 2 \cr}$$

Explanation:-

$$\eqalign{ & {{\text{3}}^{2x - y}}{\text{ = }}{{\text{3}}^{x + y}}{\text{ = }}\sqrt {{3^3}} = {3^{\frac{3}{2}}} \cr & \Leftrightarrow 2x - y = \frac{3}{2}and \cr & \,\,\,\,\,\,\,\,\,x + y = \frac{3}{2} \cr & \Leftrightarrow 3x = \frac{3}{2} + \frac{3}{2} = 3 \cr & \Leftrightarrow x = 1 \cr & \therefore y = \left( {\frac{3}{2} - 1} \right) = \frac{1}{2} \cr}$$

Explanation:-

$$\eqalign{ & {{\text{3}}^{x - y}} = 27 = {3^3} \cr & \Leftrightarrow x - y = 3........(i) \cr & {3^{x + y}} = 243 = {3^5} \cr & \Leftrightarrow x + y = 5........(ii) \cr & {\text{On solving (i) and (ii) ,}} \cr & {\text{we get }}x = 4 \cr}$$