C-04-Q230mediumsingle_mcqlog39=?\log_{\sqrt{3}} 9 = ?log39=?a222b333c444d12\dfrac{1}{2}21ব্যাখ্যাUsing logba=lnalnb\log_{b} a = \dfrac{\ln a}{\ln b}logba=lnblna: log39=log32log31/2=21/2=4\log_{\sqrt3} 9 = \dfrac{\log 3^2}{\log 3^{1/2}} = \dfrac{2}{1/2} = 4log39=log31/2log32=1/22=4.