Question Number 32960 by mondodotto@gmail.com last updated on 08/Apr/18
$$\:\boldsymbol{\mathrm{log}}_{\mathrm{2}} \boldsymbol{{x}}+\boldsymbol{\mathrm{log}}_{\mathrm{4}} \boldsymbol{{x}}+\boldsymbol{\mathrm{log}}_{\mathrm{16}} \boldsymbol{{x}}=\mathrm{7} \\ $$
Answered by Joel578 last updated on 08/Apr/18
$$\mathrm{log}_{{a}^{{n}} } \:{b}^{{m}} \:=\:\frac{{m}}{{n}}\:\mathrm{log}_{{a}} \:{b} \\ $$$$ \\ $$$$\:\:\:\:\:\:\mathrm{log}_{\mathrm{2}} \:{x}\:+\:\mathrm{log}_{\mathrm{2}^{\mathrm{2}} } \:{x}\:+\:\mathrm{log}_{\mathrm{2}^{\mathrm{4}} } \:{x}\:=\:\mathrm{7} \\ $$$$\mathrm{log}_{\mathrm{2}\:} {x}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{log}_{\mathrm{2}} \:{x}\:+\:\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{log}_{\mathrm{2}} \:{x}\:=\:\mathrm{7} \\ $$$$\frac{\mathrm{7}}{\mathrm{4}}\:\mathrm{log}_{\mathrm{2}} \:{x}\:=\:\mathrm{7}\: \\ $$$$\:\:\:\:\:\:\mathrm{log}_{\mathrm{2}} \:{x}\:=\:\mathrm{4} \\ $$$${x}\:=\:\mathrm{2}^{\mathrm{4}} \:=\:\mathrm{16} \\ $$