We are given two isosceles triangles with side lengths $x, x, a$ and $x, x, b$, where $a \neq b$, and they have equal areas. We need to find the value of $x$.

### Step-by-Step Solution

1. Area of an Isosceles Triangle:

For an isosceles triangle with sides $x, x, a$, the area $A_1$ can be calculated using the formula for the area of a triangle:

$$A_1 = \frac{1}{4} \sqrt{(x^2 + x^2 - a^2)(a^2 + 4x^2)}$$

Simplifying the term inside the square root,

$$A_1 = \frac{1}{4} \sqrt{(2x^2 - a^2)(a^2 + 4x^2)}$$

Similarly, the area $A_2$ for the triangle with sides $x, x, b$ is:

$$A_2 = \frac{1}{4} \sqrt{(2x^2 - b^2)(b^2 + 4x^2)}$$

2. Equating the Areas:

Since the areas of the two triangles are equal, we have:

$$\frac{1}{4} \sqrt{(2x^2 - a^2)(a^2 + 4x^2)} = \frac{1}{4} \sqrt{(2x^2 - b^2)(b^2 + 4x^2)}$$

Squaring both sides to remove the square root and multiplying by 16,

$$(2x^2 - a^2)(a^2 + 4x^2) = (2x^2 - b^2)(b^2 + 4x^2)$$

3. Expanding and Simplifying:

Expanding both sides of the equation gives:

$$2x^2(a^2 + 4x^2) - a^2(a^2 + 4x^2) = 2x^2(b^2 + 4x^2) - b^2(b^2 + 4x^2)$$

Simplifying the terms:

$$2x^2a^2 + 8x^4 - a^4 - 4a^2x^2 = 2x^2b^2 + 8x^4 - b^4 - 4b^2x^2$$

Combining like terms:

$$(2x^2a^2 - 4a^2x^2 - a^4) = (2x^2b^2 - 4b^2x^2 - b^4)$$

Rearranging gives:

$$(a^2 - b^2)(2x^2 - a^2 - b^2) = 0$$

4. **Solving for $x$:**

Since $a \neq b$, $a^2 - b^2 \neq 0$. Therefore,

$$2x^2 = a^2 + b^2$$

Solving for $x$,

$$x^2 = \frac{a^2 + b^2}{2}$$

Hence, the solution for $x$ is:

$$x = \frac{\sqrt{a^2 + b^2}}{\sqrt{2}} = \frac{\sqrt{a^2 + b^2}}{2}$$

Thus, the value of $x$ that satisfies the condition is:
$$\boxed{\frac{\sqrt{a^2 + b^2}}{2}}$$