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4

Part of 1972 Swedish Mathematical Competition

Problems(1)

log inequalities

Source: 1972 Swedish Mathematical Competition p4

3/26/2021
Put x=log⁡102x = \log_{10} 2, y=log⁡103y = \log_{10} 3. Then 15<1615 < 16 implies 1−x+y<4x1 - x + y < 4x, so 1+y<5x1 + y < 5x. Derive similar inequalities from 80<8180 < 81 and 243<250243 < 250. Hence show that 0.47<log⁡103<0.482. 0.47 < \log_{10} 3 < 0.482.
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