Solve this algeabric equation involving logarithms

$\begin{aligned} \color{#3D99F6} \log_{10}(x-2) + \log_{10} x & = \color{#D61F06} 1 & \small \color{#3D99F6} \text{Since }\log a + \log b = \log (ab) \\ \color{#3D99F6} \log_{10}((x-2)x) & = \color{#D61F06} \log_{10}(10^1) & \small \color{#D61F06} \text{and }\log_b (b^a) = a \\ \implies x^2 - 2x & = 10 \\ x^2 - 2x - 10 & = 0 & \small \color{#3D99F6} \text{Solve the quadratic for }x \\ x & = \frac {2+\sqrt{4+40}}2 & \small \color{#3D99F6} \text{Note that }x > 0 \text{ for }\log x \text{ to be real.} \\ & = 1 + \sqrt{11} \\ & \approx \boxed{4.317} \end{aligned}$

Since log(a) + log(b) = log(ab), log(x - 2) + log(x) = log(x(x - 2) = log(x^2 - 2x) = 1. But log(10) = 1, so x^2 - 2x - 10 = 0, with solution x = 1 + sqrt(11). Ed Gray

Log (X -2) + Log(X) = 1

Rewrite it as Log X * ( X-2) = Log (10)

The equation can be written as X^2 - 2X - 10 = 0.

X cannot be negative so the only possible solution is when X = ( 1 + SQRT(11))/2 . This evaluates to 4.316624