Question

Write log3 6 as a logarithm of base 2.

Answer 1

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