The equation you've provided is:

$8 = \sqrt{2^x}$

This can be rewritten in terms of exponents. Since $\sqrt{2^x}$ is the same as $(2^x)^{1/2}$, the equation becomes:

$8 = (2^x)^{1/2}$

Now, square both sides of the equation to eliminate the square root:

$8^2 = 2^x$

$64 = 2^x$

Since 64 is a power of 2, we can express it as:

$64 = 2^6$

Thus, we have:

$2^x = 2^6$

From this, we can conclude:

$x = 6$

Final answer:

$x = 6$

Do you want further clarification or details on any steps?

Here are 5 related questions you might explore:

1. How would the solution change if the equation was $8 = \sqrt{3^x}$?
2. What happens if you modify the equation to $8 = \sqrt{2^{x+1}}$?
3. How can you solve equations involving other roots, such as cube roots?
4. What are the rules for manipulating exponents when the bases are different?
5. How can you solve exponential equations using logarithms?

Tip: Remember, the square root of any number $a$ is the same as $a^{1/2}$. This makes it easier to manipulate exponents!