The equation you provided is:

$\left( \frac{1}{2} \right)^x = \left( 2^4 \right)^{\frac{1}{8}}$

Step-by-Step Solution:

1. Simplify the right-hand side:

   • First, simplify $2^4$: $2^4 = 16$
   • Now apply the fractional exponent: $16^{\frac{1}{8}} = (2^4)^{\frac{1}{8}} = 2^{\frac{4}{8}} = 2^{\frac{1}{2}} = \sqrt{2}$

2. Now the equation becomes: $\left( \frac{1}{2} \right)^x = \sqrt{2}$

3. Rewrite the left-hand side in terms of powers of 2: $\left( \frac{1}{2} \right)^x = (2^{-1})^x = 2^{-x}$

   So now the equation is: $2^{-x} = 2^{\frac{1}{2}}$

4. Since the bases are the same, equate the exponents: $-x = \frac{1}{2}$

5. Solve for $x$: $x = -\frac{1}{2}$

Final Answer:

$x = -\frac{1}{2}$

Would you like more details on any of the steps? Here are 5 related questions to consider:

1. How do you handle fractional exponents with different bases?
2. Can you explain how to solve exponential equations with different powers?
3. What happens when you have negative exponents in equations?
4. How do you apply logarithms to solve exponential equations?
5. What is the relationship between square roots and exponents?

Tip: When solving exponential equations, always try to express both sides of the equation using the same base—it simplifies the solution significantly!