Surds and Indices - 04

Explanation:-

$$\eqalign{ & {\left( {256} \right)^{\frac{5}{4}}} \cr & = {\left( {{4^4}} \right)^{\frac{5}{4}}} \cr & = {4^{\left( {4 \times \frac{5}{4}} \right)}} \cr & = {4^5} \cr & = 1024 \cr}$$

Explanation:-

$$\eqalign{ & {\text{Let ,}} \cr & {\left( 6 \right)^4} \div {\left( {36} \right)^3} \times 216 = {6^{\left( {x - 5} \right)}} \cr & {\text{Then,}} \cr & {6^{\left( {x - 5} \right)}} = {\left( 6 \right)^4} \div {\left( {{6^2}} \right)^3} \times {6^3} \cr & \Rightarrow {6^{\left( {x - 5} \right)}} = {6^4} \div {6^{\left( {2 \times 3} \right)}} \times {6^3} \cr & \Rightarrow {6^{\left( {x - 5} \right)}} = {6^4} \div {6^6} \times {6^3} \cr & \Rightarrow {6^{\left( {x - 5} \right)}} = {6^{\left( {4 - 6 + 3} \right)}} \cr & \Rightarrow {6^{\left( {x - 5} \right)}} = 6 \cr & \Rightarrow x - 5 = 1 \cr & \Rightarrow x = 6 \cr}$$

Explanation:-

$$\eqalign{ & {\text{Let}}\,{15.9^3} \times {\left( {81} \right)^2} \div {\left( {27} \right)^3} = {\left( 3 \right)^x}{\text{then}} \cr & \Rightarrow {\left( 3 \right)^x}{\text{ = }}\frac{{{{\left( {{3^2}} \right)}^3} \times {{\left( {{3^4}} \right)}^2}}}{{{{\left( {{3^3}} \right)}^3}}} \cr & \Rightarrow {\left( 3 \right)^x} = \frac{{{3^{\left( {2 \times 3} \right)}} \times {3^{\left( {4 \times 2} \right)}}}}{{{3^{\left( {3 \times 3} \right)}}}} \cr & \Rightarrow {\left( 3 \right)^x} = \frac{{{3^6} \times {3^8}}}{{{3^9}}} \cr & \Rightarrow {\left( 3 \right)^x} = \frac{{{3^{\left( {6 + 8} \right)}}}}{{{3^9}}} \cr & \Rightarrow {\left( 3 \right)^x} = \frac{{{3^{14}}}}{{{3^9}}} \cr & \Rightarrow {\left( 3 \right)^x} = {3^{\left( {14 - 9} \right)}} \cr & \Rightarrow {\left( 3 \right)^x} = {3^5} \cr & \Rightarrow {\left( 3 \right)^x} = 5 \cr}$$

Explanation:-

$$\eqalign{ & \frac{{16 \times 32}}{{9 \times 27 \times 81}} \cr & = \frac{{{2^4} \times {2^5}}}{{{3^2} \times {3^3} \times {3^4}}} \cr & = \frac{{{2^{\left( {4 + 5} \right)}}}}{{{3^{\left( {2 + 3 + 4} \right)}}}} \cr & = \frac{{{2^9}}}{{{3^9}}} \cr & = {\left( {\frac{2}{3}} \right)^9} \cr}$$

Explanation:-

$$\eqalign{ & \frac{{343 \times 49}}{{216 \times 16 \times 81}} \cr & = \frac{{{7^3} \times {7^2}}}{{{6^3} \times {2^4} \times {3^4}}} \cr & = \frac{{{7^{\left( {3 + 2} \right)}}}}{{{6^3} \times {{\left( {2 \times 3} \right)}^4}}} \cr & = \frac{{{7^5}}}{{{6^3} \times {6^4}}} \cr & = \frac{{{7^5}}}{{{6^{\left( {3 + 4} \right)}}}} \cr & = \frac{{{7^5}}}{{{6^7}}} \cr}$$

Explanation:-

3⁸ × 3⁴
= 3^{(8 + 4)}
= 3¹²
= (3⁶)²
= (729)²

Explanation:-

$$\eqalign{ & {{\text{3}}^{10}} \times {\text{2}}{{\text{7}}^2}{\text{ = }}{{\text{9}}^2} \times {{\text{3}}^n} \cr & \Rightarrow {{\text{3}}^{10}} \times {\left( {{3^3}} \right)^2}{\text{ = }}{\left( {{3^2}} \right)^2} \times {{\text{3}}^n} \cr & \Rightarrow {{\text{3}}^{10 + 6}}{\text{ = }}{{\text{3}}^{4 + n}} \cr & \Rightarrow 10 + 6 = 4 + n \cr & \Rightarrow n = 12 \cr}$$

Explanation:-

$$\eqalign{ & \frac{{{{\left( {1.5} \right)}^3} + {{\left( {4.7} \right)}^3} + {{\left( {3.8} \right)}^3} - 3 \times 1.5 \times 4.7 \times 3.8}}{{{{\left( {1.5} \right)}^2} + {{\left( {4.7} \right)}^2} + {{\left( {3.8} \right)}^2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5}} \cr & = \frac{{\left\{ {{{\left( {1.5} \right)}^2} + {{\left( {4.7} \right)}^2} + {{\left( {3.8} \right)}^2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5} \right\}}}{{\left\{ {{{\left( {1.5} \right)}^2} + {{\left( {4.7} \right)}^2} + {{\left( {3.8} \right)}^2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5} \right\}}} \cr & ....\left[ {\therefore {a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)} \right] \cr & = 1.5 + 4.7 + 3.8 \cr & = 10.0 \cr & = 10 \cr}$$

Explanation:-

$$\eqalign{ & \frac{{{{\left( {1000} \right)}^{12}}}}{{{{\left( {10} \right)}^{30}}}} \cr & = \frac{{{{\left( {{{10}^3}} \right)}^{12}}}}{{{{\left( {10} \right)}^{30}}}} \cr & = \frac{{{{\left( {10} \right)}^{\left( {3 \times 12} \right)}}}}{{{{\left( {10} \right)}^{30}}}} \cr & = \frac{{{{\left( {10} \right)}^{36}}}}{{{{\left( {10} \right)}^{30}}}} \cr & = {\left( {10} \right)^{\left( {36 - 30} \right)}} \cr & = {10^6} \cr & = {\left( {{{10}^3}} \right)^2} \cr & = {\left( {1000} \right)^2} \cr}$$

Explanation:-

$$\eqalign{ & \frac{1}{{\sqrt {\left( {12 - \sqrt {140} } \right)} }} - \frac{1}{{\sqrt {\left( {8 - \sqrt {60} } \right)} }} - \frac{2}{{\sqrt {\left( {10 + \sqrt {84} } \right)} }} \cr & = \frac{1}{{\sqrt {\left( {12 - \sqrt {4 \times 35} } \right)} }} - \frac{1}{{\sqrt {\left( {8 - \sqrt {4 \times 15} } \right)} }} - \frac{2}{{\sqrt {\left( {10 + \sqrt {4 \times 21} } \right)} }} \cr & = \frac{1}{{\sqrt {\left( {12 - 2\sqrt {35} } \right)} }} - \frac{1}{{\sqrt {\left( {8 - 2\sqrt {15} } \right)} }} - \frac{2}{{\sqrt {\left( {10 - 2\sqrt {21} } \right)} }} \cr & = \frac{1}{{\sqrt {{{\left( {\sqrt 7 } \right)}^2} + {{\left( {\sqrt 5 } \right)}^2} - 2.\sqrt 7 .\sqrt 5 } }} - \frac{1}{{\sqrt {{{\left( {\sqrt 5 } \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} - 2.\sqrt 5 .\sqrt 3 } }} - \frac{2}{{\sqrt {{{\left( {\sqrt 7 } \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} + 2.\sqrt 7 .\sqrt 3 } }} \cr & = \frac{1}{{\sqrt {{{\left( {\sqrt 7 - \sqrt 5 } \right)}^2}} }} - \frac{1}{{\sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} }} - \frac{2}{{\sqrt {{{\left( {\sqrt 7 + \sqrt 3 } \right)}^2}} }} \cr & = \frac{1}{{\sqrt 7 - \sqrt 5 }} - \frac{1}{{\sqrt 5 - \sqrt 3 }} - \frac{2}{{\sqrt 7 + \sqrt 3 }} \cr & {\text{Rationlizing in above equation,}} \cr & \frac{1}{{\sqrt 7 - \sqrt 5 }} \times \frac{{\sqrt 7 + \sqrt 5 }}{{\sqrt 7 + \sqrt 5 }} - \frac{1}{{\sqrt 5 - \sqrt 3 }} \times \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} - \frac{2}{{\sqrt 7 + \sqrt 3 }} \times \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 7 - \sqrt 3 }} \cr & = \frac{{\sqrt 7 + \sqrt 5 }}{2} - \frac{{\sqrt 5 + \sqrt 3 }}{2} - \frac{{\sqrt 7 - \sqrt 3 }}{2} \cr & = \frac{{\sqrt 7 + \sqrt 5 - \sqrt 5 - \sqrt 3 - \sqrt 7 + \sqrt 3 }}{2} \cr & = 0 \cr}$$