To solve the equation $3\log_x 2 + \log_2 x = 4$, we will use the change of base formula for logarithms, specifically $\log_b a = \frac{1}{\log_a b}$. Step 1: Rewrite the first term using the change of base formula. $$ \log_x 2 = \frac{1}{\log_2 x} $$ Substitute this into the original equation: $$ 3\left(\frac{1}{\log_2 x}\right) + \log_2 x = 4 $$ $$ \frac{3}{\log_2 x} + \log_2 x = 4 $$ Step 2: Let $y = \log_2 x$ to simplify the equation. $$ \frac{3}{y} + y = 4 $$ Step 3: Solve the equation for $y$. Multiply the entire equation by $y$ (assuming $y \neq 0$). $$ 3 + y^2 = 4y $$ Rearrange the terms to form a quadratic equation: $$ y^2 - 4y + 3 = 0 $$ Step 4: Factor the quadratic equation. 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 $y$: $$ y - 1 = 0 \implies y = 1 $$ $$ y - 3 = 0 \implies y = 3 $$ Step 5: Substitute back $y = \log_2 x$ and solve for $x$. 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, $x = 2^3$. $$ x = 8 $$ Step 6: Check the solutions. 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 $\boxed{x=2, 8}$.