To solve the equation \( \frac{k(5) \cdot k(2)}{k(x)} = k(2)^{-2} \), we can first simplify the left-hand side.
Recall that \( k(a) \) represents a power of some base (like \( k(x) = k^x \)). Thus, we have:
- \( k(5) = k^5 \)
- \( k(2) = k^2 \)
- \( k(x) = k^x \)
Now, substituting these into the equation, we get:
\[ \frac{k^5 \cdot k^2}{k^x} = k^2 \cdot k^{-2} \]
Next, simplify both sides:
Left side:
\[ \frac{k^5 \cdot k^2}{k^x} = \frac{k^{5+2}}{k^x} = \frac{k^7}{k^x} = k^{7-x} \]
Right side:
\[ k^2 \cdot k^{-2} = k^{2-2} = k^0 = 1 \]
Now we have:
\[ k^{7-x} = 1 \]
For this equation to hold true for any base \( k \) (where \( k \neq 0 \) or \( k \neq 1 \)), the exponent must be zero:
\[ 7 - x = 0 \]
Solving for \( x \):
\[ x = 7 \]
After analyzing the provided options, it looks like 7 is not listed. If you meant to include it, please check if there's an error. However, since it's not an option provided, if this is an error in the problem, I can't provide an answer based on the incorrect premise.
If you had \( x = 7 \) and if that was a choice, that would be the answer. Otherwise, if you stick with the choices you've given, none apply. Please double-check the equation or the options.
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