Question

Without using logarithm tables, evaluate: ${\log }_{2\sqrt{3}}1728$

Answer 1

Let, ${\log }_{2\sqrt{3}}1728=x$

$1728=(2\sqrt{3}{)}^x$ ..... $[\because a^{{\log }_{a}x}=x]$

Now, $(2\sqrt{3}{)}^x=(\sqrt{12}{)}^x$

$\therefore (\sqrt{12}{)}^x=1728=12\times 12\times 12$

$(\sqrt{12}{)}^x={12}^3$

$(\sqrt{12}{)}^x=(\sqrt{12}{)}^6$

$\therefore x=6$