Question Number 188569 by 073 last updated on 03/Mar/23
Answered by JDamian last updated on 03/Mar/23
$$\mathrm{16}? \\ $$
Commented by 073 last updated on 03/Mar/23
$$\mathrm{solution}?? \\ $$
Answered by mr W last updated on 03/Mar/23
$${h}\left({ad}\right)={h}\left({a}\right)+{h}\left({d}\right) \\ $$$$\Rightarrow{h}\left({x}\right)=\mathrm{log}_{{b}} \:{x} \\ $$$$\frac{{h}\left({a}^{\mathrm{4}} \right)}{{h}\left(\sqrt[{\mathrm{4}}]{{a}}\right)}=\frac{\mathrm{4}\:\mathrm{log}_{{b}} \:{a}}{\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{log}_{{b}} \:{a}}=\mathrm{16} \\ $$
Answered by manxsol last updated on 03/Mar/23
$${h}\left({a}^{\mathrm{4}} \right)={h}\left({a}^{\frac{\mathrm{1}}{\mathrm{4}}} .{a}^{\frac{\mathrm{1}}{\mathrm{4}}} ……..\right)=\mathrm{16}{h}\left({a}^{\frac{\mathrm{1}}{\mathrm{4}}} \right) \\ $$$${h}\left(^{\mathrm{4}} \sqrt{{a}}\right)={h}\left({a}^{\frac{\mathrm{1}}{\mathrm{4}}} \right) \\ $$$$\frac{{h}\left({a}^{\mathrm{4}} \right)}{{h}\left(^{\mathrm{4}} \sqrt{{a}}\right)}=\frac{\mathrm{16}{h}\left({a}^{\frac{\mathrm{1}}{\mathrm{4}}} \right)}{{h}\left({a}^{\frac{\mathrm{1}}{\mathrm{4}}} \right)}=\mathrm{16} \\ $$