Let $a$ and $b$ be real numbers greater than 1 such that $ab=100$. The maximum possible value of $a^{(\log_{10}b)^2}$ can be written in the form $10^x$ for some real number $x$. Find $x$.
1. **Express \(a\) and \(b\) in terms of \(k\):**
   Given \(ab = 100\), we can write \(a\) and \(b\) in terms of a variable \(k\) such that:
   \[
   a = 10^{2-k} \quad \text{and} \quad b = 10^k
   \]
   This ensures that:
   \[
   ab = 10^{2-k} \cdot 10^k = 10^2 = 100
   \]

2. **Express the given function in terms of \(k\):**
   We need to find the maximum value of \(a^{(\log_{10}b)^2}\). Substituting the expressions for \(a\) and \(b\):
   \[
   a^{(\log_{10}b)^2} = (10^{2-k})^{(\log_{10}(10^k))^2} = (10^{2-k})^{k^2}
   \]
   Simplifying the exponent:
   \[
   (10^{2-k})^{k^2} = 10^{k^2(2-k)}
   \]

3. **Find the maximum value of the exponent \(k^2(2-k)\):**
   We need to maximize the function \(f(k) = k^2(2-k)\) over the interval \([0, 2]\).

4. **Differentiate \(f(k)\) to find critical points:**
   \[
   f(k) = k^2(2-k)
   \]
   Differentiate \(f(k)\) with respect to \(k\):
   \[
   f'(k) = \frac{d}{dk} [k^2(2-k)] = \frac{d}{dk} [2k^2 - k^3] = 4k - 3k^2
   \]
   Set the derivative equal to zero to find critical points:
   \[
   4k - 3k^2 = 0
   \]
   Factor out \(k\):
   \[
   k(4 - 3k) = 0
   \]
   Thus, the critical points are:
   \[
   k = 0 \quad \text{or} \quad k = \frac{4}{3}
   \]

5. **Evaluate \(f(k)\) at the critical points and endpoints:**
   \[
   f(0) = 0^2(2-0) = 0
   \]
   \[
   f\left(\frac{4}{3}\right) = \left(\frac{4}{3}\right)^2 \left(2 - \frac{4}{3}\right) = \frac{16}{9} \cdot \frac{2}{3} = \frac{32}{27}
   \]
   \[
   f(2) = 2^2(2-2) = 0
   \]

6. **Determine the maximum value:**
   The maximum value of \(f(k)\) on the interval \([0, 2]\) is \(\frac{32}{27}\).

7. **Express the maximum value in the form \(10^x\):**
   Since \(a^{(\log_{10}b)^2} = 10^{k^2(2-k)}\), the maximum value is:
   \[
   10^{\frac{32}{27}}
   \]
   Therefore, \(x = \frac{32}{27}\).

The final answer is \(\boxed{\frac{32}{27}}\).