Let the function $f\left(x\right)=\left\{\begin{array}{cc}\frac{{\log }_e\left(1+5x\right)-{\log }_e\left(1+\alpha x\right)}{x}& \mathrm{if} x\ne 0\\ 10& \mathrm{if} x=0\end{array}\right.$ be continuous at $x=0$. Then $\alpha$ is equal to

$f\left(x\right)=\left\{\begin{array}{cc}\frac{\ln \left(1+5x\right)-\ln \left(1+\alpha x\right)}{x}& x\ne 0\\ 10& x=0\end{array}\right.$

For $f\left(x\right)$ to be continuous

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

$\underset{x\to 0}{\lim }\frac{\ln \left(1+5x\right)-\ln \left(1+\alpha x\right)}{x}=10$

Using expansion of $\ln \left(1+x\right)=x-\frac{x^2}{2}+\frac{x^3}{3}- ....$, we get

$\underset{x\to 0}{\lim }\frac{\left(5x-{\displaystyle \frac{25x^2}{2}}+\dots \dots \right)-\left(\alpha x-{\displaystyle \frac{{\alpha }^2{x}^2}{2}}+\dots \dots \right)}{x}=10$

i.e. $5-\alpha =10\Rightarrow \alpha =-5$