Answer: \(x=4\).

Use the change of base formula:

\(\log_x10=\dfrac{\lg10}{\lg x}=\dfrac{1}{\lg x}.\)

Therefore

\(\dfrac{2}{\log_x10}=2\lg x.\)

The equation becomes

\(2\lg x-\lg(x+4)=\lg2.\)

Since \(2\lg x=\lg(x^2)\),

\(\lg(x^2)-\lg(x+4)=\lg2.\)

Using \(\lg A-\lg B=\lg\left(\dfrac{A}{B}\right)\),

\(\lg\left(\dfrac{x^2}{x+4}\right)=\lg2.\)

So

\(\dfrac{x^2}{x+4}=2.\)

Hence

\(x^2=2x+8,\)

so

\(x^2-2x-8=0.\)

Factorise:

\((x-4)(x+2)=0.\)

The question states \(x\gt0\), so \(x=-2\) is rejected.

Therefore

\(x=4.\)