Question

Write log3 6 as a logarithm of base 2.

Answer 1

`Use:\\\log_ab=(\log_cb)/(\log_ca)\ and\ \log_ab+\log_ac=\log_a(bc)\\\\\log_36=(\log_26)/(\log_23)=(\log_2(2\cdot3))/(\log_23)=(\log_22+\log_23)/(\log_23)\\\\=(\log_22)/(\log_23)+(\log_23)/(\log_23)=\boxed{(1)/(\log_23)+1}\\\\other\ method\\Use:\\\log_ab=(1)/(\log_ba)\\\\\log_36=\log_3(2\cdot3)=\log_32+\log_33=\log_32+1=\boxed{(1)/(\log_23)+1}`