Solve the logarithmic equation: x 2x + x 54 = 5.

The given equation is $\log_x 2x + \log_x 54 = 5$.
Using the logarithm property $\log_b M + \log_b N = \log_b (MN)$, we can combine the terms on the left side:

$$\log_x (2x \cdot 54) = 5$$ $$\log_x (108x) = 5$$

The definition of a logarithm states that if $\log_b A = C$, then $b^C = A$. Applying this to our equation:

$$x^5 = 108x$$

Rearrange the equation to set it to zero:

$$x^5 - 108x = 0$$

Factor out $x$:

$$x(x^4 - 108) = 0$$

This gives two possible cases:
Case 1: $x = 0$
Case 2: $x^4 - 108 = 0 \implies x^4 = 108$

For a logarithm $\log_b A$ to be defined, the base $b$ must satisfy $b > 0$ and $b \neq 1$. Also, the argument $A$ must satisfy $A > 0$.

Case 1: If $x=0$, the base of the logarithm would be 0, which is not allowed. Also, the term $2x$ would be 0, which is not allowed as an argument. Therefore, $x=0$ is not a valid solution.

Case 2: If $x^4 = 108$, then $x = \pm \sqrt[4]{108}$.
Since the base $x$ must be positive, we take the positive root:

$$x = \sqrt[4]{108}$$

Let's check if this value satisfies the conditions for the base:

• $x = \sqrt[4]{108}$ is clearly greater than 0.
• $x = \sqrt[4]{108}$ is not equal to 1 (since $1^4 = 1$ and $2^4 = 16$, $3^4 = 81$, $4^4 = 256$, $\sqrt[4]{108}$ is between 3 and 4). The arguments $2x$ and 54 will both be positive if $x$ is positive. Thus, $x = \sqrt[4]{108}$ is the only valid solution.