Question

Prove that $\frac{2}{5}<lo{g}_{10}3<\frac{1}{2}$

Answer 1

$2lo{g}_{10}3=lo{g}_{10}9<1\therefore lo{g}_{10}3<\frac{1}{2}$
$\frac{2}{5}<lo{g}_{10}3$ if 2 < 5 $lo{g}_{10}3=lo{g}_{10}{3}^5$
or $lo{g}_{10}$243 > 2. It is true because
$lo{g}_{10}$243 > $lo{g}_{10}{10}^2=2$