It is given that $\frac{2^x \times 32(2^x)}{8^{x+1}} = \frac{9(5^{2x}) - 5^{2x+1}}{5^x - 5^{x-1}}$. Evaluate $10^x$ without using a calculator. Solve $\sqrt{x+7} - x - 1 = 0$.

1. Simplify the left-hand side (LHS) of the equation by expressing all terms in base 2.

$\text{LHS} = \frac{2^x \times 2^5 \times 2^x}{(2^3)^{x+1}}$
$= \frac{2^{2x+5}}{2^{3x+3}}$
$= 2^{(2x+5) - (3x+3)}$
$= 2^{-x+2}$

2. Simplify the right-hand side (RHS) of the equation by factoring out common terms.

$\text{RHS} = \frac{9(5^{2x}) - 5(5^{2x})}{5^x - \frac{1}{5}(5^x)}$
$= \frac{4(5^{2x})}{\frac{4}{5}(5^x)}$
$= 5(5^{2x - x})$
$= 5(5^x) = 5^{x+1}$

3. Equate the simplified LHS and RHS, and rearrange to find $10^x$.

$2^{-x+2} = 5^{x+1}$
$2^2 \times 2^{-x} = 5^1 \times 5^x$
$4 \times \frac{1}{2^x} = 5 \times 5^x$
$\frac{4}{5} = 2^x \times 5^x$
$10^x = \frac{4}{5}$

1. Isolate the surd and square both sides

$\sqrt{x+7} = x + 1$
$x + 7 = (x + 1)^2$

2. Expand and form a quadratic equation

$x + 7 = x^2 + 2x + 1$
$x^2 + x - 6 = 0$

3. Solve the quadratic equation

$(x + 3)(x - 2) = 0$
$x = -3 \text{ or } x = 2$

4. Check for extraneous roots and state the final answer

Substitute $x = -3$ into the original equation:
$\text{LHS} = \sqrt{-3+7} - (-3) - 1 = \sqrt{4} + 3 - 1 = 4 \neq 0$
So $x = -3$ is rejected.

Substitute $x = 2$ into the original equation:
$\text{LHS} = \sqrt{2+7} - 2 - 1 = \sqrt{9} - 3 = 0$
So $x = 2$ is the valid solution.