Question Number 197125 by Amidip last updated on 08/Sep/23
Commented by som(math1967) last updated on 08/Sep/23
$$\boldsymbol{{b}}=? \\ $$
Answered by Frix last updated on 09/Sep/23
$${a}=\frac{\mathrm{ln}\:\mathrm{18}}{\mathrm{ln}\:\mathrm{12}}=\frac{\mathrm{2ln}\:\mathrm{3}\:+\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}\:+\mathrm{2ln}\:\mathrm{2}}=\frac{\mathrm{2}{p}+{q}}{{p}+\mathrm{2}{q}} \\ $$$${b}=\frac{\mathrm{ln}\:\mathrm{54}}{\mathrm{ln}\:\mathrm{24}}=\frac{\mathrm{3ln}\:\mathrm{3}\:+\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}\:+\mathrm{3ln}\:\mathrm{2}}=\frac{\mathrm{3}{p}+{q}}{{p}+\mathrm{3}{q}} \\ $$$${ab}+\mathrm{5}\left({a}−{b}\right)= \\ $$$$=\frac{\mathrm{6}{p}^{\mathrm{2}} +\mathrm{5}{pq}+{q}^{\mathrm{2}} }{\left({p}+\mathrm{2}{q}\right)\left({p}+\mathrm{3}{q}\right)}−\frac{\mathrm{5}\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} \right)}{\left({p}+\mathrm{2}{q}\right)\left({p}+\mathrm{3}{q}\right)}= \\ $$$$=\frac{{p}^{\mathrm{2}} +\mathrm{5}{pq}+\mathrm{6}{q}^{\mathrm{2}} }{\left({p}+\mathrm{2}{q}\right)\left({p}+\mathrm{3}{q}\right)}=\frac{\left({p}+\mathrm{2}{q}\right)\left({p}+\mathrm{3}{q}\right)}{\left({p}+\mathrm{2}{q}\right)\left({p}+\mathrm{3}{q}\right)}=\mathrm{1} \\ $$