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MISCELLANEOUS EXERCISE 6 (II) · Q177

Q.Without using log tables, prove that 25<log⁡103<12\dfrac25<\log_{10}3<\dfrac12.

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Upper bound: 32=9<10  ⟹  3<10=101/2  ⟹  log⁡103<log⁡10101/2=123^2=9<10 \implies 3<\sqrt{10}=10^{1/2} \implies \log_{10}3<\log_{10}10^{1/2}=\dfrac12 (since log⁡10\log_{10} is increasing).

Lower bound: we want to show log⁡103>25\log_{10}3>\dfrac25, i.e. 3>102/53>10^{2/5}, i.e. (raising both sides to the 5th power, which preserves the inequality since both sides are positive) 35>1023^5>10^2. …

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