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(D) Given the equation: $x^{1 + \log_{10} x} = 100000x$Taking $\log_{10}$ on both sides:$(1 + \log_{10} x) \cdot \log_{10} x = \log_{10} (100000x)$Let

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(D) Given the equation: $x^{1 + \log_{10} x} = 100000x$Taking $\log_{10}$ on both sides:$(1 + \log_{10} x) \cdot \log_{10} x = \log_{10} (100000x)$Let $y = \log_{10} x$. Then the equation becomes:$(1 + y)y = \log_{10} (10^5) + \log_{10} x$$y + y^2 = 5 + y$$y^2 = 5$$y = \pm \sqrt{5}$Since $y = \log_{10} x$,we have $\log_{10} x = \sqrt{5}$ or $\log_{10} x = -\sqrt{5}$.Thus,$x_1 = 10^{\sqrt{5}}$ and $x_2 = 10^{-\sqrt{5}}$.The product of the solutions is $x_1 \cdot x_2 = 10^{\sqrt{5}} \cdot 10^{-\sqrt{5}} = 10^{\sqrt{5} - \sqrt{5}} = 10^0 = 1$.

The product of all the solutions of the equation $x^{1 + \log_{10} x} = 100000x$ is

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