Question Number 193360 by MATHEMATICSAM last updated on 11/Jun/23
$$\mathrm{If}\:{x}\:=\:\mathrm{2}^{{p}} \:\mathrm{and}\:{y}\:=\:\mathrm{4}^{{q}} \:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{log}_{\mathrm{2}} \left({x}^{\mathrm{3}} {y}\right)\:=\:\mathrm{3}{p}\:+\:\mathrm{2}{q} \\ $$
Answered by AST last updated on 11/Jun/23
$${log}_{\mathrm{2}} \left(\mathrm{2}^{\mathrm{3}{p}} \mathrm{2}^{\mathrm{2}{q}} \right)={log}_{\mathrm{2}} \left(\mathrm{2}^{\mathrm{3}{p}+\mathrm{2}{q}} \right)=\mathrm{3}{p}+\mathrm{2}{q} \\ $$
Answered by Frix last updated on 11/Jun/23
$$\left({b}^{{m}} \right)^{{n}} ={b}^{{mn}} \\ $$$${b}^{{m}} {b}^{{n}} ={b}^{{m}+{n}} \\ $$$$\mathrm{log}_{{b}} \:{b}^{{a}} \:={a} \\ $$$$ \\ $$$${x}^{\mathrm{3}} {y}=\mathrm{2}^{\mathrm{3}{p}} \mathrm{4}^{{q}} =\mathrm{2}^{\mathrm{3}{p}} \left(\mathrm{2}^{\mathrm{2}} \right)^{{q}} =\mathrm{2}^{\mathrm{3}{p}} \mathrm{2}^{\mathrm{2}{q}} =\mathrm{2}^{\mathrm{3}{p}+\mathrm{2}{q}} \\ $$$$\mathrm{log}_{\mathrm{2}} \:\mathrm{2}^{\mathrm{3}{p}+\mathrm{2}{q}} \:=\mathrm{3}{p}+\mathrm{2}{q} \\ $$