If $$\alpha$$ and $$\beta(\alpha > \beta)$$ satisfy the equation $$x^{1 + \log_{10} x} = 10x$$ then the value of $$\alpha + \frac{1}{\beta}$$ is equal to

Applying logarithms to base 10 on both sides we get

$$\log_{10}{x^{1 + \log_{10} x} }= \log_{10}{10x}$$

$$({1 + \log_{10} x})\log_{10}{x }= \log_{10}{10} + \log_{10}{x} $$

$$({1 + \log_{10} x})\log_{10}{x }= 1 + \log_{10}{x} $$

writing $$t = \log_{10} x$$, the equation becomes

$$(1 + t)t = 1 + t$$, 

$$t^2 = 1$$ 

$$t = \pm 1$$. 

$$log_{10}{x} = \pm 1$$

$$x = 10, x = \dfrac{1}{10}$$

Since we know that $$\alpha>\beta$$, the roots are $$\alpha = 10$$ and $$\beta = \dfrac{1}{10}$$, 

This gives us

$$\alpha + \dfrac{1}{\beta} = 10 + 10 = 20$$.