Kris is asked to compute $\log_{10} (x^y)$, where $y$ is a positive integer and $x$ is a positive real number. However, they misread this as $(\log_{10} x)^y$ , and compute this value. Despite the reading error, Kris still got the right answer. Given that $x > 10^{1.5}$ , determine the largest possible value of $y$.
1. We start with the given problem: Kris is asked to compute $\log_{10} (x^y)$, but instead computes $(\log_{10} x)^y$. Despite the error, Kris gets the correct answer. We need to determine the largest possible value of $y$ given that $x > 10^{1.5}$.

2. Let $a = x^y$. Then, $\log_{10} (x^y) = \log_{10} (a)$. By the properties of logarithms, we have:
   \[
   \log_{10} (x^y) = y \log_{10} (x)
   \]

3. Kris computes $(\log_{10} x)^y$ instead. For Kris to get the correct answer despite the error, we must have:
   \[
   (\log_{10} x)^y = y \log_{10} (x)
   \]

4. Let $b = \log_{10} (x)$. Then, the equation becomes:
   \[
   b^y = yb
   \]

5. Dividing both sides by $b$ (since $b \neq 0$), we get:
   \[
   b^{y-1} = y
   \]

6. We know that $x > 10^{1.5}$, so:
   \[
   \log_{10} (x) > 1.5
   \]
   Therefore, $b > 1.5$.

7. We need to find the largest integer $y$ such that:
   \[
   b^{y-1} = y
   \]
   and $b > 1.5$.

8. Let's test values of $y$ to find the largest possible value:
   - For $y = 2$:
     \[
     b^{2-1} = b = 2 \quad \text{(not possible since $b > 1.5$)}
     \]
   - For $y = 3$:
     \[
     b^{3-1} = b^2 = 3 \quad \Rightarrow \quad b = \sqrt{3} \approx 1.732 \quad \text{(possible since $b > 1.5$)}
     \]
   - For $y = 4$:
     \[
     b^{4-1} = b^3 = 4 \quad \Rightarrow \quad b = \sqrt[3]{4} \approx 1.587 \quad \text{(possible since $b > 1.5$)}
     \]
   - For $y = 5$:
     \[
     b^{5-1} = b^4 = 5 \quad \Rightarrow \quad b = \sqrt[4]{5} \approx 1.495 \quad \text{(not possible since $b \leq 1.5$)}
     \]

9. From the above calculations, the largest possible value of $y$ such that $b > 1.5$ is $y = 4$.

The final answer is $\boxed{4}$