We are tasked with solving the equation:

$\frac{\sqrt{\sqrt{\sqrt{\sqrt{5 + 2^x}}}}}{2^x - 1} = \frac{1}{3^{7/8}}$

Step-by-step solution:

1. Simplify both sides: Start by rewriting the equation:

   $\frac{\left(5 + 2^x\right)^{1/16}}{2^x - 1} = \frac{1}{3^{7/8}}$

   The left-hand side (LHS) is $(5 + 2^x)^{1/16}$ because the square root is applied four times (which is equivalent to taking the 16th root of the expression).

2. Raise both sides to the 16th power: To eliminate the fractional exponents on the left-hand side, raise both sides of the equation to the 16th power:

   $\left( \frac{(5 + 2^x)^{1/16}}{2^x - 1} \right)^{16} = \left( \frac{1}{3^{7/8}} \right)^{16}$

   This simplifies to:

   $\frac{5 + 2^x}{(2^x - 1)^{16}} = \frac{1}{3^{14}}$

3. Cross-multiply:

   $5 + 2^x = \frac{(2^x - 1)^{16}}{3^{14}}$

   Now we have a more complex equation. Solving this exactly will involve checking for values of $x$ that make both sides equal.

Trial and Error for $x$:

Let's try small values for $x$:

• For $x = 1$:

  $5 + 2^1 = 5 + 2 = 7$ The right-hand side becomes: $\frac{(2^1 - 1)^{16}}{3^{14}} = \frac{(2 - 1)^{16}}{3^{14}} = \frac{1^{16}}{3^{14}} = \frac{1}{3^{14}}$ Clearly, $7 \neq \frac{1}{3^{14}}$, so $x = 1$ is not a solution.

• For $x = 2$:

  $5 + 2^2 = 5 + 4 = 9$ The right-hand side becomes: $\frac{(2^2 - 1)^{16}}{3^{14}} = \frac{(4 - 1)^{16}}{3^{14}} = \frac{3^{16}}{3^{14}} = 3^2 = 9$ This is a match.

Thus, $x = 2$ is the solution.

Final answer:

$x = 2$

Would you like further details or have any questions?

Here are five related questions for further exploration:

1. How would the solution change if the exponent on the denominator was different?
2. What happens if we replace $5 + 2^x$ with $3 + 2^x$?
3. Can you generalize the method used to solve similar nested radical equations?
4. What would happen if the right-hand side was $\frac{1}{4^{7/8}}$?
5. How would you graph the function $y = \frac{\sqrt{\sqrt{\sqrt{\sqrt{5 + 2^x}}}}}{2^x - 1}$?

Tip: Always try small integer values for $x$ when solving such exponential equations—it can lead to the solution quickly!