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Q.15. f (x) = ${\log }_{100x}\left(\frac{2{\log }_{10}x+1}{-x}\right)$ exists, if ..... .

Dear Student,

$$\begin{gathered} p\left(x\right) = {\log }_{a}x exists if a > 0, x > 0 and a \ne 1 \\ So, for f\left(x\right) = {\log }_{100x}\left(\frac{2{\log }_{10}x + 1}{-x}\right), \\ 100x > 0, \frac{2{\log }_{10}x + 1}{-x} > 0 and 100x \ne 1 \\ Now, 100x > 0 \Rightarrow x > 0 \\ and, 100x \ne 1 \Rightarrow x \ne \frac{1}{100} \\ For \frac{2{\log }_{10}x + 1}{-x} > 0, there are two cases possible: \\ Case 1: 2{\log }_{10}x + 1 > 0 and -x > 0 \\ \Rightarrow {\log }_{10}x > \frac{-1}{2} and x < 0, which is not possible as for {\log }_{10}x to exist, x has to be greater than 0 \\ Case 2: 2{\log }_{10}x + 1 < 0 and -x < 0 \\ \Rightarrow {\log }_{10}x < \frac{-1}{2} and x > 0 \\ \Rightarrow x < {10}^{\frac{-1}{2}} and x > 0 \\ \Rightarrow 0 < x < \frac{1}{\sqrt{10}} \\ Combining the three conditions, we see that f\left(x\right) exists if 0 < x < \frac{1}{\sqrt{10}} and x \ne \frac{1}{100}. \end{gathered}$$​

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