To solve for \( x \) in the equation \( 110_x = 40_5 \), we need to convert both numbers to a common base. We'll use base 10 (decimal) for this conversion.

### Step 1: Convert \( 40_5 \) to decimal.
The number \( 40_5 \) is in base 5. To convert it to base 10:
\[ 40_5 = 4 \times 5^1 + 0 \times 5^0 = 4 \times 5 = 20 \]

So, \( 40_5 = 20_{10} \).

### Step 2: Express \( 110_x \) in decimal.
The number \( 110_x \) in base \( x \) is expressed as:
\[ 110_x = 1 \times x^2 + 1 \times x^1 + 0 \times x^0 = x^2 + x \]

### Step 3: Set up the equation.
We know that:
\[ x^2 + x = 20 \]

### Step 4: Solve the quadratic equation.
The equation \( x^2 + x - 20 = 0 \) can be solved using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1 \), \( b = 1 \), and \( c = -20 \).

Plugging in the values, we get:
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-20)}}{2 \cdot 1} \]
\[ x = \frac{-1 \pm \sqrt{1 + 80}}{2} \]
\[ x = \frac{-1 \pm \sqrt{81}}{2} \]
\[ x = \frac{-1 \pm 9}{2} \]

This gives us two solutions:
\[ x = \frac{-1 + 9}{2} = \frac{8}{2} = 4 \]
\[ x = \frac{-1 - 9}{2} = \frac{-10}{2} = -5 \]

Since \( x \) represents a base and must be a positive integer, we discard \( x = -5 \). Therefore, the valid solution is:
\[ x = 4 \]

Would you like more details or have any questions?

Here are 5 questions you might consider next:
1. How do you convert numbers from other bases to decimal?
2. How do you solve a quadratic equation?
3. What are the properties of different number bases?
4. Can you explain how the quadratic formula is derived?
5. How do you determine the validity of a solution in number base problems?

**Tip:** When converting numbers from one base to another, always express both numbers in a common base for easier comparison and solving.

Learn how to find x in the base conversion problem 110_x = 40_5 using decimal conversion and quadratic equation solving. This problem explores number bases and quadratic equations, suitable for high school level students.

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