3log2 to the base x + logx to the base 2=4

$$\log_x 2 = \frac{1}{\log_2 x}$$

Substitute this into the original equation:

$$3\left(\frac{1}{\log_2 x}\right) + \log_2 x = 4$$ $$\frac{3}{\log_2 x} + \log_2 x = 4$$

Multiply the entire equation by $y$ (assuming $y \neq 0$).

$$3 + y^2 = 4y$$

Rearrange the terms to form a quadratic equation:

$$y^2 - 4y + 3 = 0$$

We need two numbers that multiply to 3 and add to -4. These numbers are -1 and -3.

$$(y - 1)(y - 3) = 0$$

This gives two possible values for $y$:

$$y - 1 = 0 \implies y = 1$$ $$y - 3 = 0 \implies y = 3$$

Case 1: $y = 1$

$$\log_2 x = 1$$

By the definition of a logarithm, $x = 2^1$.

$$x = 2$$

Case 2: $y = 3$

$$\log_2 x = 3$$

By the definition of a logarithm, $x = 2^3$.

$$x = 8$$

For $x=2$:
$3\log_2 2 + \log_2 2 = 3(1) + 1 = 3+1 = 4$. This is correct.

For $x=8$:
$3\log_8 2 + \log_2 8 = 3\left(\frac{1}{3}\right) + 3 = 1+3 = 4$. This is correct.

Both solutions are valid.

The solutions are $x=2, 8$.