Question Number 137764 by byaw last updated on 06/Apr/21
$$\mathrm{If}\:\mathrm{log}_{\mathrm{2}} \mathrm{3}={a}\:\mathrm{and}\:\mathrm{log}_{\mathrm{3}} \mathrm{7}={b},\:\mathrm{express} \\ $$$$\mathrm{log}_{\mathrm{42}} \mathrm{56}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{a}\:\mathrm{and}\:{b} \\ $$
Answered by EnterUsername last updated on 06/Apr/21
$${log}_{\mathrm{2}} \mathrm{3}={a},\:{log}_{\mathrm{3}} \mathrm{7}=\frac{{log}_{\mathrm{2}} \mathrm{7}}{{log}_{\mathrm{2}} \mathrm{3}}={b}\Rightarrow{log}_{\mathrm{2}} \mathrm{7}={ab} \\ $$$${log}_{\mathrm{42}} \mathrm{56}=\frac{{log}_{\mathrm{2}} \mathrm{56}}{{log}_{\mathrm{2}} \mathrm{42}}=\frac{{log}_{\mathrm{2}} \left(\mathrm{8}×\mathrm{7}\right)}{{log}_{\mathrm{2}} \left(\mathrm{6}×\mathrm{7}\right)}=\frac{{log}_{\mathrm{2}} \mathrm{2}^{\mathrm{3}} +{log}_{\mathrm{2}} \mathrm{7}}{{log}_{\mathrm{2}} \mathrm{2}+{log}_{\mathrm{2}} \mathrm{3}+{log}_{\mathrm{2}} \mathrm{7}} \\ $$$$=\frac{\mathrm{3}+{ab}}{\mathrm{1}+{a}+{ab}} \\ $$
Commented by otchereabdullai@gmail.com last updated on 08/Apr/21
$$\mathrm{wow}! \\ $$