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.
1 answer