Indices and Surds Questions in English

(B) Given equation: $12^{4+2x^2} = (24\sqrt{3})^{3x^2-2}$
Express $24\sqrt{3}$ in terms of base $12$:
$24\sqrt{3} = 12 \times 2 \times \sqrt{3} = 12 \times \sqrt{4} \times \sqrt{3} = 12 \times \sqrt{12} = 12^1 \times 12^{1/2} = 12^{3/2}$
Substituting this into the equation:
$12^{4+2x^2} = (12^{3/2})^{3x^2-2}$
$12^{4+2x^2} = 12^{\frac{3}{2}(3x^2-2)}$
Equating the exponents:
$4+2x^2 = \frac{3}{2}(3x^2-2)$
$8+4x^2 = 9x^2-6$
$9x^2-4x^2 = 8+6$
$5x^2 = 14$
$x^2 = \frac{14}{5}$
$x = \pm \sqrt{\frac{14}{5}}$

(B) Given the equation: $20^{2-3x^2} = (40\sqrt{5})^{3x^2-2}$.
Notice that $40\sqrt{5} = 20 \times 2\sqrt{5} = 20 \times \sqrt{4 \times 5} = 20 \times \sqrt{20} = 20^1 \times 20^{1/2} = 20^{3/2}$.
Substituting this into the equation,we get: $20^{2-3x^2} = (20^{3/2})^{3x^2-2}$.
Equating the exponents: $2-3x^2 = \frac{3}{2}(3x^2-2)$.
Let $y = 3x^2-2$. Then the equation becomes $-y = \frac{3}{2}y$.
This implies $y(1 + \frac{3}{2}) = 0$,so $y = 0$.
Thus,$3x^2-2 = 0$,which gives $3x^2 = 2$.
Therefore,$x^2 = \frac{2}{3}$,so $x = \pm \sqrt{\frac{2}{3}}$.

(C) Given the equation: $4^{x} - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2x - 1}$
Rearranging terms: $4^{x} + 2^{2x - 1} = 3^{x + \frac{1}{2}} + 3^{x - \frac{1}{2}}$
Since $2^{2x - 1} = \frac{4^{x}}{2}$,we have: $4^{x} + \frac{4^{x}}{2} = 3^{x} \cdot \sqrt{3} + \frac{3^{x}}{\sqrt{3}}$
$4^{x} \left(1 + \frac{1}{2}\right) = 3^{x} \left(\frac{3 + 1}{\sqrt{3}}\right)$
$4^{x} \left(\frac{3}{2}\right) = 3^{x} \left(\frac{4}{\sqrt{3}}\right)$
$\left(\frac{4}{3}\right)^{x} = \frac{4}{\sqrt{3}} \cdot \frac{2}{3} = \frac{8}{3\sqrt{3}}$
$\left(\frac{4}{3}\right)^{x} = \left(\frac{2}{\sqrt{3}}\right)^{3} = \left(\left(\frac{2}{\sqrt{3}}\right)^{2}\right)^{\frac{3}{2}} = \left(\frac{4}{3}\right)^{\frac{3}{2}}$
Comparing exponents,we get $x = \frac{3}{2}$.

(C) Given equation: $4^x - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2x - 1}$
Rearranging terms: $2^{2x} + 2^{2x - 1} = 3^{x + \frac{1}{2}} + 3^{x - \frac{1}{2}}$
Factor out common terms: $2^{2x}(1 + \frac{1}{2}) = 3^x(\sqrt{3} + \frac{1}{\sqrt{3}})$
Simplify: $2^{2x}(\frac{3}{2}) = 3^x(\frac{3 + 1}{\sqrt{3}}) = 3^x(\frac{4}{\sqrt{3}})$
Multiply both sides by $\sqrt{3}$: $2^{2x} \cdot 3 \cdot \sqrt{3} = 3^x \cdot 8$
Express in powers: $2^{2x} \cdot 3^{\frac{3}{2}} = 3^x \cdot 2^3$
Equating powers of $2$: $2x = 3 \Rightarrow x = \frac{3}{2}$
Equating powers of $3$: $x = \frac{3}{2}$
Thus,$x = \frac{3}{2}$.

(B) Given the equation: $20^{2-3x^2} = (40\sqrt{5})^{3x^2-2}$.
Note that $40\sqrt{5} = 20 \times 2 \times \sqrt{5} = 20 \times \sqrt{4} \times \sqrt{5} = 20 \times \sqrt{20} = 20^{1} \times 20^{1/2} = 20^{3/2}$.
Substituting this into the equation:
$20^{2-3x^2} = (20^{3/2})^{3x^2-2}$
$20^{2-3x^2} = 20^{\frac{3}{2}(3x^2-2)}$
Since the bases are equal,we equate the exponents:
$2-3x^2 = \frac{3}{2}(3x^2-2)$
$2-3x^2 = -\frac{3}{2}(2-3x^2)$
$(2-3x^2) + \frac{3}{2}(2-3x^2) = 0$
$(2-3x^2)(1 + \frac{3}{2}) = 0$
$(2-3x^2)(\frac{5}{2}) = 0$
Since $\frac{5}{2} \neq 0$,we must have $2-3x^2 = 0$.
$3x^2 = 2$
$x^2 = \frac{2}{3}$
$x = \pm \sqrt{\frac{2}{3}}$.

(B) We simplify the expression from the innermost radical outward:
$\sqrt{14-6 \sqrt{5}} = \sqrt{9+5-2(3)(\sqrt{5})} = \sqrt{(3-\sqrt{5})^2} = 3-\sqrt{5}$.
Substituting this back into the expression:
$\sqrt{6-3 \sqrt{5} + (3-\sqrt{5})} = \sqrt{9-4 \sqrt{5}}$.
We simplify $\sqrt{9-4 \sqrt{5}}$ as $\sqrt{9-2(2)(\sqrt{5})} = \sqrt{(\sqrt{5}-2)^2} = \sqrt{5}-2$.
Now,substitute this into the main expression:
$\sqrt{2+\sqrt{5}-(\sqrt{5}-2)} = \sqrt{2+\sqrt{5}-\sqrt{5}+2} = \sqrt{4} = 2$.

(C) Let $x = \frac{\sqrt{8+\sqrt{28}}+\sqrt{8-\sqrt{28}}}{\sqrt{8+\sqrt{28}}-\sqrt{8-\sqrt{28}}}$.
Rationalizing the denominator:
$x = \frac{(\sqrt{8+\sqrt{28}}+\sqrt{8-\sqrt{28}})^2}{(\sqrt{8+\sqrt{28}})^2 - (\sqrt{8-\sqrt{28}})^2}$
$x = \frac{(8+\sqrt{28}) + (8-\sqrt{28}) + 2\sqrt{(8+\sqrt{28})(8-\sqrt{28})}}{(8+\sqrt{28}) - (8-\sqrt{28})}$
$x = \frac{16 + 2\sqrt{64-28}}{2\sqrt{28}}$
$x = \frac{16 + 2\sqrt{36}}{2\sqrt{4 \times 7}}$
$x = \frac{16 + 2(6)}{2(2\sqrt{7})}$
$x = \frac{16 + 12}{4\sqrt{7}} = \frac{28}{4\sqrt{7}} = \frac{7}{\sqrt{7}} = \sqrt{7}$.

(D) To compare the numbers,we express them with a common exponent by finding the least common multiple $(LCM)$ of the indices $3$ and $4$,which is $12$.
The numbers are:
$4^{1/3} = (4^4)^{1/12} = 256^{1/12}$
$5^{1/4} = (5^3)^{1/12} = 125^{1/12}$
$7^{1/4} = (7^3)^{1/12} = 343^{1/12}$
$8^{1/3} = (8^4)^{1/12} = 4096^{1/12}$
Comparing the bases $256, 125, 343, 4096$,the smallest value is $125$.
Therefore,$125^{1/12} = \sqrt[4]{5}$ is the least number.