$$\eqalign{ & {\text{Given}}\,{\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{b}{a}} \right)^{x - 3}} \cr & \Rightarrow {\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{a}{b}} \right)^{ - \left( {x - 3} \right)}} = {\left( {\frac{a}{b}} \right)^{\left( {3 - x} \right)}} \cr & \Rightarrow x - 1 = 3 - x \cr & \Rightarrow 2x = 4 \cr & \Rightarrow x = 2 \cr} $$

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

$$\eqalign{ & \frac{{{4^3} \times {5^4}}}{{{4^5}}} \cr & = \frac{{{5^4}}}{{{4^{\left( {5 - 3} \right)}}}} \cr & = \frac{{{5^4}}}{{{4^2}}} \cr & = \frac{{625}}{{16}} \cr & = 39.0625 \cr} $$

$$\eqalign{ & {\left( {0.04} \right)^{ - 1.5}} = {\left( {\frac{4}{{100}}} \right)^{ - 1.5}} \cr & = {\left( {\frac{1}{{25}}} \right)^{ - \left( {3/2} \right)}} \cr & = {\left( {25} \right)^{\left( {3/2} \right)}} \cr & = {\left( {{5^2}} \right)^{\left( {3/2} \right)}} \cr & = {\left( 5 \right)^{2 \times \left( {3/2} \right)}} \cr & = {5^3} \cr & = 125 \cr} $$

$$\eqalign{ & \frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} \cr & = \frac{1}{{\left( {1 + \frac{{{a^n}}}{{{a^m}}}} \right)}} + \frac{1}{{\left( {1 + \frac{{{a^m}}}{{{a^n}}}} \right)}} \cr & = \frac{{{a^m}}}{{\left( {{a^m} + {a^n}} \right)}} + \frac{{{a^n}}}{{\left( {{a^m} + {a^n}} \right)}} \cr & = \frac{{\left( {{a^m} + {a^n}} \right)}}{{\left( {{a^m} + {a^n}} \right)}} \cr & = 1 \cr} $$

We know that 11² = 121.
Putting m = 11 and n = 2, we get:
(m - 1)^{n + 1} = (11 - 1)^{(2 + 1)} = 10³ = 1000

$$\eqalign{ & {x^z} = {y^2} \Leftrightarrow {10^{\left( {0.48z} \right)}} = {10^{2 \times 0.70}} = {10^{1.40}} \cr & \Rightarrow 0.48z = 1.40 \cr & \Rightarrow z = \frac{{140}}{{48}} = \frac{{35}}{{12}} = 2.9({\text{approx}}) \cr} $$

5^a = 3125     ⇔     5^a = 5⁵
⇒ a = 5.
∴ 5^{(a - 3)} = 5^{(5 - 3)} = 5² = 25

3^{x - y} = 27 = 3³     ⇔     x - y = 3 ....(i)
3^{x + y} = 243 = 3⁵     ⇔     x + y = 5 ....(ii)
On solving (i) and (ii), we get x = 4.

$$\eqalign{ & {\text{Given Exp}}{\text{.}} \cr & = {x^{\left( {b - c} \right)\left( {b + c - a} \right)}}.{x^{\left( {c - a} \right)\left( {c + a - b} \right)}}.{x^{\left( {a - b} \right)\left( {a + b - c} \right)}} \cr} $$
  $$ = {x^{\left( {b - c} \right)\left( {b + c} \right) - a\left( {b - c} \right)}}.$$    $${x^{\left( {c - a} \right)\left( {c + a} \right) - b\left( {c - a} \right)}}.$$   $${x^{\left( {a - b} \right)\left( {a + b} \right) - c\left( {a - b} \right)}}$$
$$\eqalign{ & = {x^{\left( {{b^2} - {c^2} + {c^2} - {a^2} + {a^2} - {b^2}} \right)}}.{x^{ - a\left( {b - c} \right) - b\left( {c - a} \right) - c\left( {a - b} \right)}} \cr & = {{x^0} . {x^\left( -ab + ac - bc + ba - ca + cb \right)}} \cr & = \left( {{x^0} \times {x^0}} \right) \cr & = \left( {1 \times 1} \right) \cr & = 1 \cr} $$

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