Question

a^n = 5 + (n = 1) x 2

x y
2 ?
4 ?

Answer 1

The given equation appears to involve a relationship between \( a^n \), and from the way it is written, it seems to indicate a pattern based on the values of \( n \).

Let's interpret the equation \( a^n = 5 + 2 \times (n = 1) \):

1. The expression \( (n = 1) \) evaluates to 1 when \( n \) is 1, and 0 otherwise.
2. Therefore, when \( n = 1 \):
\[
a^1 = 5 + 2 \times 1 = 5 + 2 = 7 \quad \Rightarrow \quad a = 7
\]
3. For \( n = 2 \):
\[
a^2 = 5 + 2 \times 0 = 5 + 0 = 5 \quad \Rightarrow \quad a^2 = 5 \quad \Rightarrow \quad a = \sqrt{5}
\]
4. For other values of \( n \), the term \( (n = 1) \) would be 0, hence:
\[
a^n = 5
\]
for \( n = 3, 4, 5, \ldots \) which gives \( a = \sqrt[n]{5} \).

Now, to solve for \( x \) and \( y \):

- If \( x = 2 \), that means \( n = 2 \):
\[
a^2 = 5 \quad \Rightarrow \quad y = 5
\]

- If \( x = 4 \), that means \( n = 4 \):
\[
a^4 = 5 \quad \Rightarrow \quad y = 5
\]

Therefore, we can summarize the results as follows:

| x | y |
|---|----|
| 2 | 5 |
| 4 | 5 |

So for the values provided:
- When \( x = 2 \), \( y = 5 \)
- When \( x = 4 \), \( y = 5 \)