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GoldenRose99
@GoldenRose99
July 2022
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логарифм 1/125 по основанию 0,2
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hlopushinairina
Log₀,₂(1/125)=log₁/₅(1/5³)=log₅⁻¹(5⁻³)=(-3)·(-1)log₅5=3;
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