Solve the following equation with a rational exponent: 2x32=16 .(1 point)

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**.