Question

$\sf{}Prove \: \: that \: \: \: log_{10} \: \: 2 \: \: lies \: \: between \: \\ \sf{}\: \frac{1}{3} and \: \frac{1}{4}$
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Answer 1

Answer:

$\huge \fbox \color{lime}{꧁Solution࿐}$

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To prove:-

$\tt \: \frac{1}{3} < log_{10}(2) < \frac{1}{4}$

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Proof:-

Step1- Simplification of term

$\tt {2}^{12} = 4096$

=> 1000 < 4096 < 10000

$\tt = > {10}^{3} < {2}^{12} < {10}^{4}$

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Step2- Taking logarithm to the base 10

$\tt{10}^{3} < {2}^{12} < {10}^{4}$

$\tt log_{10}( {10}^{3} ) < log_{10}( {2}^{12} ) < log_{10}( {10}^{4} )$

$\tt = > 3 \: log_{10}(10) < 12 \: log_{10}(2) < 4 \: log_{10}(10)$

$\tt = > 3 < 12 \: log_{10}(2) < 4$

$\tt = > \frac{3}{12} < log_{10}(2) < \frac{4}{12}$

$\tt \green{= > \frac{1}{3} < log_{10}(2) < \frac{1}{4} }$

Hence proved

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Answer 2

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