If $f\left(x\right)=\frac{\log \left(1+2x\right)-\log \left(1-3x\right)}{x}$ is continuous at $x=0,$ then $f\left(0\right)=$

$f\left(0\right)=\underset{x\to 0}{Lt}f\left(x\right)$ 

        = $\underset{x\to 0}{\lim }\frac{\log (1+2x)-\log (1-3x)}{x}$ 

        =$\underset{x\to 0}{\lim }\frac{1og(1+2x)}{x}-\underset{x\to 0}{\lim }\frac{1og(1-3x)}{x}$

       $=2–\left(-3\right)$

      $f\left(0\right)=5$