Question 12
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
Solution
Taking logarithms to base 10 and writing $$t = \log_{10} x$$, the equation becomes $$(1 + t)t = 1 + t$$, so $$t^2 = 1$$ and $$t = \pm 1$$. Hence the roots are $$\alpha = 10$$ and $$\beta = \frac{1}{10}$$, giving $$\alpha + \frac{1}{\beta} = 10 + 10 = 20$$.
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