log(x-1) / log(x−1) = 2 log(x-2)

The equation shown in the image is $\log(x-1) \div \log(x-1) = 2 \log(x-2)$.
For the logarithms to be defined, their arguments must be positive:

• For $\log(x-1)$, we need $x-1 > 0 \implies x > 1$.
• For $\log(x-2)$, we need $x-2 > 0 \implies x > 2$. Additionally, the denominator $\log(x-1)$ cannot be zero. This means $x-1 \neq 1$, so $x \neq 2$. Combining these conditions, the valid domain for $x$ is $x > 2$.

Since $x > 2$, $\log(x-1)$ is defined and non-zero. Any non-zero quantity divided by itself is 1.

So the equation simplifies to:

Divide both sides by 2:

Assuming the base of the logarithm is 10 (common logarithm, as no base is specified):

Add 2 to both sides:

The approximate value of $\sqrt{10}$ is 3.162.
So, $x \approx 2 + 3.162 = 5.162$.
This value satisfies the domain condition $x > 2$. Therefore, the solution is valid.