The problem asks to solve the logarithmic equation $\log_2{x} + \log_x{4} = 3$.

1. Problem Description

The problem asks to solve the logarithmic equation $\log_2{x} + \log_x{4} = 3$.

2. Solution Steps

We are given the equation $\log_2{x} + \log_x{4} = 3$.

Using the change of base formula, we can rewrite $\log_x{4}$ as $\frac{\log_2{4}}{\log_2{x}}$. Since $\log_2{4} = 2$, the equation becomes:

$\log_2{x} + \frac{2}{\log_2{x}} = 3$

Let $y = \log_2{x}$. Then the equation becomes:

$y + \frac{2}{y} = 3$

Multiplying both sides by $y$, we get:

$y^2 + 2 = 3y$

$y^2 - 3y + 2 = 0$

Factoring the quadratic equation, we have:

$(y-1)(y-2) = 0$

So, $y = 1$ or $y = 2$.

If $y = 1$, then $\log_2{x} = 1$, which means $x = 2^1 = 2$.

If $y = 2$, then $\log_2{x} = 2$, which means $x = 2^2 = 4$.

Therefore, the solutions are $x=2$ and $x=4$.

3. Final Answer

The solutions are 2 or

4.