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To solve the equation \( 2x^{32} = 16 \), we start by isolating the term with the variable.
First, we can rewrite 16 as a power of 2:
\[
16 = 2^4
\]
So, the equation becomes:
\[
2x^{32} = 2^4
\]
Next, we can divide both sides by 2:
\[
x^{32} = \frac{2^4}{2} = 2^{4-1} = 2^3
\]
Now we have:
\[
x^{32} = 2^3
\]
We can take the \( 32^{nd} \) root of both sides or raise both sides to the power of \( \frac{1}{32} \):
\[
x = (2^3)^{\frac{1}{32}}
\]
Using the rule of exponents \( (a^m)^n = a^{mn} \):
\[
x = 2^{\frac{3}{32}}
\]
This value of \( x \) is a positive real number, and since it is not one of the provided integer options (84, no solution, 4, or 8), the correct answer is:
**no solution**.