Question

show that log 10 root 2 lies between 1 by 12 and 1 by 6​

Answer 1

$\frac{1}{12} = 0.0833.... \\ and \\ \frac{1}{6} = 0.1666....$

now,

$log( \sqrt{2} ) = \frac{ log(2) }{2} (becoz. \: log( {a}^{m}) = m log(a) ) \\ = \frac{0.30103}{2} = 0.150515 \\ and \: we \: know \: that \\ 0.0833... < 0.150515 < 0.1666... \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: < hence \: proved >$