For what positive integers n>1 does the expression 10n−1log102⋅log103⋯log10n attain the smallest possible value? Find this value.
Solution
Let us compare the expressions 10n−2log102log103⋯log10(n−1) and 10n−1log102log103⋯log10n. The inequality 10n−2log102log103⋯log10(n−1)≥10n−1log102log103⋯log10n holds if and only if 1≥101log10n=log1010n, which is equivalent to 10≥10n and 1010≥n. This implies 10log102>102log102⋅log103>⋯>101010−2log102⋅log103⋯log10(1010−1)=101010−1log102⋅log103⋯log10(1010) and 101010−1log102⋅log103⋯log10(1010)<<101010log102⋅log103⋯log10(1010+1)101010+1log102⋅log103⋯log10(1010+2)<… So, the expression 10n−1log102log103⋯log10n has the smallest possible value when n=1010−1 or n=1010. For these two values of n the value of the expression is 101010−1log102log103⋯log101010
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Source: MathNet,
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