Let $a$ be a positive real number, $n$ a positive integer, and define the [i]power tower[/i] $a\uparrow n$ recursively with $a\uparrow 1=a$, and $a\uparrow(i+1)=a^{a\uparrow i}$ for $i=1,2,3,\ldots$.  For example, we have $4\uparrow 3=4^{(4^4)}=4^{256}$, a number which has $155$ digits.  For each positive integer $k$, let $x_k$ denote the unique positive real number solution of the equation $x\uparrow k=10\uparrow (k+1)$.  Which is larger: $x_{42}$ or $x_{43}$?
1. We start by defining the power tower \( a \uparrow n \) recursively:
   \[
   a \uparrow 1 = a
   \]
   \[
   a \uparrow (i+1) = a^{a \uparrow i} \quad \text{for} \quad i = 1, 2, 3, \ldots
   \]
   For example, \( 4 \uparrow 3 = 4^{(4^4)} = 4^{256} \).

2. We are given the equation \( x \uparrow k = 10 \uparrow (k+1) \) and need to determine which is larger: \( x_{42} \) or \( x_{43} \).

3. Notice that \( x_n > 10 \) for all \( n \). We will show that the sequence \( x_n \) is strictly decreasing.

4. Consider the equation \( x_n \uparrow k = 10 \uparrow (k+1) \). Taking the natural logarithm of both sides, we get:
   \[
   \ln(x_n \uparrow k) = \ln(10 \uparrow (k+1))
   \]
   Using the properties of logarithms, we can rewrite this as:
   \[
   (x_n \uparrow (k-1)) \ln x_n = (10 \uparrow k) \ln 10
   \]

5. Similarly, for \( x_{n-1} \), we have:
   \[
   (x_{n-1} \uparrow (k-1)) \ln x_{n-1} = (10 \uparrow k) \ln 10
   \]

6. Since \( x_n \uparrow (k-1) = x_{n-1} \uparrow (k-1) \), we can equate the two expressions:
   \[
   (x_n \uparrow (k-1)) \ln x_n = (x_{n-1} \uparrow (k-1)) \ln x_{n-1}
   \]

7. Dividing both sides by \( x_n \uparrow (k-1) \) and \( x_{n-1} \uparrow (k-1) \), we get:
   \[
   \ln x_n = \ln x_{n-1}
   \]

8. Since \( x_n > 10 \) and \( x_{n-1} > 10 \), we have:
   \[
   \frac{x_n}{x_{n-1}} < \frac{\ln 10}{\ln x_n} < 1
   \]
   This implies that \( x_n < x_{n-1} \) for all \( n \).

9. Specifically, \( x_{42} \) is larger than \( x_{43} \).

The final answer is \( \boxed{ x_{42} } \) is larger than \( x_{43} \).