Surds And Indices

$${\text{1}}{{\text{0}}^{ - 8x}}$$
$$ = {\left[ {{{10}^x}} \right]^{ - 8}}$$
$$ = {\left[ {\frac{1}{2}} \right]^{ - 8}}$$
$$ = {2^8}$$
$$ = 256$$

$$\left( {\frac{1}{{1.4}} + \frac{1}{{4.7}} + \frac{1}{{7.10}} + \frac{1}{{10.13}} + \frac{1}{{13.16}}} \right)$$
Formula :
$$\frac{1}{{{\text{Difference denominator value}}}} \times $$     $$\left[ {\frac{1}{{{\text{First value}}}} - \frac{1}{{{\text{Last value}}}}} \right]$$
$$ = \frac{1}{3} \times $$ $$\left[ {1 - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + \frac{1}{7} - \frac{1}{{10}} + \frac{1}{{10}} - \frac{1}{{13}} + \frac{1}{{13}} - \frac{1}{{16}}} \right]$$
$$ = \frac{1}{3} \times \left[ {1 - \frac{1}{{16}}} \right]$$
$$ = \frac{1}{3} \times \frac{{15}}{{16}}$$
$$ = \frac{5}{{16}}$$

$$\frac{1}{{\sqrt 5 + \sqrt 3 }} \times \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }}$$
$$ = \frac{{\sqrt 5 - \sqrt 3 }}{{5 - 3}}$$
$$ = \frac{{2.236 - 1.732}}{2}$$
$$ = \frac{{0.504}}{2}$$
$$ = 0.252$$

$${\text{ }}{\left( {\frac{3}{5}} \right)^3}{\left( {\frac{3}{5}} \right)^{ - 6}} = {\left( {\frac{3}{5}} \right)^{2x - 1}}$$
$$ ⇒ {\text{ }}{\left( {\frac{3}{5}} \right)^{\left( {3 - 6} \right)}} = {\left( {\frac{3}{5}} \right)^{2x - 1}}$$
$$ ⇒ {\text{ }}{\left( {\frac{3}{5}} \right)^{ - 3}} = {\left( {\frac{3}{5}} \right)^{2x - 1}}$$
$$ ⇒ 2x - 1 = - 3$$
$$ ⇒ 2x = - 2$$
$$ ⇒ x = - 1$$

$${\text{5}}\sqrt 5 \times {5^3} \div {5^{ - \frac{3}{2}}}{\text{ = }}{{\text{5}}^{a + 2}}$$
$$ ⇒ \frac{{5 \times {5^{\frac{1}{2}}} \times {5^3}}}{{{5^{ - \frac{3}{2}}}}} = {5^{a + 2}}$$
$$ ⇒ {5^{\left( {1 + \frac{1}{2} + 3 + \frac{3}{2}} \right)}} = {5^{a + 2}}$$
$$ ⇒ {5^6} = {5^{a + 2}}$$
$$ ⇒ a + 2 = 6$$
$$ ⇒ a = 4$$

$${{\text{2}}^x} = \root 3 \of {32} $$
$$ ⇒ {{\text{2}}^x} = {\left( {32} \right)^{\frac{1}{3}}} = {\left( {{2^5}} \right)^{\frac{1}{3}}}$$
$$ ⇒ x = \frac{5}{3}$$

$$\sqrt {12 + \sqrt {12 + \sqrt {12 + ..... } } } $$
(3, 4) are the factor of 4
If there is '+' in $$\sqrt {\,\,\,} $$  Answer is Highest value
If there is '-' in $$\sqrt {\,\,\,} $$  Answer is lowest value.
So the answer is 4

$${{\text{5}}^{\left( {x + 3} \right)}}{\text{ = 2}}{{\text{5}}^{(3x - 4)}}$$
$$ ⇒ {{\text{5}}^{\left( {x + 3} \right)}}{\text{ = }}{\left( {{5^2}} \right)^{(3x - 4)}}$$
$$ ⇒ {{\text{5}}^{\left( {x + 3} \right)}}{\text{ = }}{{\text{5}}^2}^{(3x - 4)}$$
$$ ⇒ {{\text{5}}^{\left( {x + 3} \right)}}{\text{ = }}{{\text{5}}^{(6x - 8)}}$$
$$ ⇒ x + 3 = 6x - 8$$
$$ ⇒ 5x = 11$$
$$ ⇒ x = \frac{{11}}{5}$$

$$\frac{{{2^{n + 4}} - 2\left( {{2^n}} \right)}}{{2\left( {{2^{n + 3}}} \right)}}{\text{ }}$$
$$ = \frac{{{2^{n + 4}} - {2^{n + 1}}}}{{{2^{n + 4}}}}{\text{ }}$$
$$ = \frac{{{2^{n + 4}}}}{{{2^{n + 4}}}} - \frac{{{2^{n + 1}}}}{{{2^{n + 4}}}}{\text{ }}$$
$$ = 1 - {2^{(n + 1) - \left( {n + 4} \right)}}$$
$${\text{ = 1}} - {2^{ - 3}}$$
$${\text{ = 1}} - \frac{1}{8}{\text{ }}$$
$${\text{ = }}\frac{7}{8}$$

$$a + b = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} + \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}$$
$$ = \frac{{{{\left( {\sqrt 5 + 1} \right)}^2} + {{\left( {\sqrt 5 - 1} \right)}^2}}}{{\left( {\sqrt 5 - 1} \right)\left( {\sqrt 5 + 1} \right)}}$$
$$ = \frac{{2\left[ {{{\left( {\sqrt 5 } \right)}^2} + 1} \right]}}{{5 - 1}}$$
$$ = \frac{{2\left( {5 + 1} \right)}}{4}$$
$$ = 3$$
$$a.b = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} \times \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} = 1$$
$${\text{Put value in expression}}$$
$$\frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}$$
$$ = \frac{{{{\left( {a + b} \right)}^2} - ab}}{{{{\left( {a + b} \right)}^2} - 3ab}}$$
$$ = \frac{{{3^2} - 1}}{{{3^2} - 3}}$$
$$ = \frac{{9 - 1}}{{9 - 3}}$$
$$ = \frac{4}{3}$$