$$\begin{aligned} \text{If }x = \left(8 + 3\sqrt{7}\right),\text{ what is the value of }\\\left(\sqrt{x} - \dfrac{1}{\sqrt{x}}\right)? \end{aligned}$$

Answer: Option C

Explanation:

$$\begin{aligned} 6^{m-2}\\ = \dfrac{6^m}{6^2}\\ = \dfrac{46656}{6^2}\\ = \dfrac{46656}{36} = 1296 \end{aligned}$$

Answer: Option B

Explanation:

$$\begin{aligned} = 256^{0.16+0.09} = 256^{0.25} = 256^{\frac{25}{100}} \end{aligned}$$

$$\begin{aligned} = 256^{\frac{1}{4}}= (4^4)^{\frac{1}{4}} \end{aligned}$$

$$\begin{aligned} =(4)^{4 \times \frac{1}{4}} = 4 \end{aligned}$$

Answer: Option D

Explanation:

Lets assume,
$$\begin{aligned} \left(25 \right)^{7.5} \times \left(5 \right)^{2.5} \div \left(125 \right)^{1.5} = 5^x \\ \text{then, } \frac{ \left( 5^2 \right)^{7.5} \times \left(5 \right)^{2.5} }{\left(5^3 \right)^{1.5} } = 5^x \\ => \text{then, } \frac{ \left( 5^{15} \right) \times \left(5^{2.5} \right) }{\left(5^{4.5} \right) } = 5^x \\ => 5^x = 5^{15 + 2.5 - 4.5} \\ => 5^x = 5^{13} \\ \text{Hence, } x = 13 \end{aligned}$$

Answer: Option D

Explanation:

We know that $$\begin{aligned} (11)^2 = 121 \end{aligned}$$
So, putting values in said equation we get,
$$\begin{aligned} (11-1)^{2 + 1} = (10)^3 = 1000 \end{aligned}$$

Answer: Option B

Explanation:

$$\begin{align}&\left(\sqrt{x} - \dfrac{1}{\sqrt{x}}\right)^2\\\\ &= x - 2 + \dfrac{1}{x}\\\\ &= x + \dfrac{1}{x} - 2 \\\\ &= \left(8 + 3\sqrt{7}\right) + \dfrac{1}{\left(8 + 3\sqrt{7}\right)} - 2 \\\\ &= \left(8 + 3\sqrt{7}\right) + \dfrac{\left(8 - 3\sqrt{7}\right)}{\left(8 + 3\sqrt{7}\right)\left(8 - 3\sqrt{7}\right)} - 2 \\\\ &= \left(8 + 3\sqrt{7}\right) + \dfrac{\left(8 - 3\sqrt{7}\right)}{8^2 - \left(3\sqrt{7}\right)^2} - 2 \\\\ &= \left(8 + 3\sqrt{7}\right) + \dfrac{\left(8 - 3\sqrt{7}\right)}{64 - 63} - 2 \\\\ &= \left(8 + 3\sqrt{7}\right) + \dfrac{\left(8 - 3\sqrt{7}\right)}{1} - 2 \\\\ &= 8 + 3\sqrt{7} + 8 - 3\sqrt{7} - 2 \\\\ &= 14 \\\\ &\text{as }\left(\sqrt{x} - \dfrac{1}{\sqrt{x}}\right)^2 = 14\\\\ &\text{so ,}\left(\sqrt{x} - \dfrac{1}{\sqrt{x}}\right) = \sqrt{14}\end{align}$$