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Question Number 130242 by Raxreedoroid last updated on 23/Jan/21

$$\mathrm{Find}m$$$$n={\left(p_1{p}_2\right)}^m$$$$2n=\left(p_1^{1+{log}_{p_1{p}_2}n}\right)\left(p_2^{1+{log}_{p_1{p}_2}n}\right)$$

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