To solve the equation \( 142_n = 47_{10} \), we first need to convert \( 142_n \) (a number in base \( n \)) into base 10.
The value of \( 142_n \) in decimal form is computed as follows:
\[ 142_n = 1 \cdot n^2 + 4 \cdot n^1 + 2 \cdot n^0 \]
This simplifies to:
\[ 142_n = 1 \cdot n^2 + 4 \cdot n + 2 \]
So we need to set this equal to \( 47_{10} \):
\[ n^2 + 4n + 2 = 47 \]
Next, we move \( 47 \) over to the left side of the equation:
\[ n^2 + 4n + 2 - 47 = 0 \]
This simplifies to:
\[ n^2 + 4n - 45 = 0 \]
Now we can solve this quadratic equation using the quadratic formula, which is:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 4 \), and \( c = -45 \). Substituting these values into the formula:
\[ n = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-45)}}{2 \cdot 1} \] \[ n = \frac{-4 \pm \sqrt{16 + 180}}{2} \] \[ n = \frac{-4 \pm \sqrt{196}}{2} \] \[ n = \frac{-4 \pm 14}{2} \]
Now we calculate the two possible values for \( n \):
- \( n = \frac{10}{2} = 5 \)
- \( n = \frac{-18}{2} = -9 \) (not a valid base)
Since base \( n \) must be a positive integer greater than 4 (because of the digit 4 in \( 142_n \)), we conclude:
\[ n = 5 \]
Thus, the solution to the equation \( 142_n = 47_{10} \) is:
\[ \boxed{5} \]
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