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Question Number 197125 by Amidip last updated on 08/Sep/23

Commented by som(math1967) last updated on 08/Sep/23

b=?

$$\boldsymbol{{b}}=? \\ $$

Answered by Frix last updated on 09/Sep/23

a=((ln 18)/(ln 12))=((2ln 3 +ln 2)/(ln 3 +2ln 2))=((2p+q)/(p+2q)) b=((ln 54)/(ln 24))=((3ln 3 +ln 2)/(ln 3 +3ln 2))=((3p+q)/(p+3q)) ab+5(a−b)= =((6p^2 +5pq+q^2 )/((p+2q)(p+3q)))−((5(p^2 −q^2 ))/((p+2q)(p+3q)))= =((p^2 +5pq+6q^2 )/((p+2q)(p+3q)))=(((p+2q)(p+3q))/((p+2q)(p+3q)))=1

$${a}=\frac{\mathrm{ln}\:\mathrm{18}}{\mathrm{ln}\:\mathrm{12}}=\frac{\mathrm{2ln}\:\mathrm{3}\:+\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}\:+\mathrm{2ln}\:\mathrm{2}}=\frac{\mathrm{2}{p}+{q}}{{p}+\mathrm{2}{q}} \\ $$$${b}=\frac{\mathrm{ln}\:\mathrm{54}}{\mathrm{ln}\:\mathrm{24}}=\frac{\mathrm{3ln}\:\mathrm{3}\:+\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}\:+\mathrm{3ln}\:\mathrm{2}}=\frac{\mathrm{3}{p}+{q}}{{p}+\mathrm{3}{q}} \\ $$$${ab}+\mathrm{5}\left({a}−{b}\right)= \\ $$$$=\frac{\mathrm{6}{p}^{\mathrm{2}} +\mathrm{5}{pq}+{q}^{\mathrm{2}} }{\left({p}+\mathrm{2}{q}\right)\left({p}+\mathrm{3}{q}\right)}−\frac{\mathrm{5}\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} \right)}{\left({p}+\mathrm{2}{q}\right)\left({p}+\mathrm{3}{q}\right)}= \\ $$$$=\frac{{p}^{\mathrm{2}} +\mathrm{5}{pq}+\mathrm{6}{q}^{\mathrm{2}} }{\left({p}+\mathrm{2}{q}\right)\left({p}+\mathrm{3}{q}\right)}=\frac{\left({p}+\mathrm{2}{q}\right)\left({p}+\mathrm{3}{q}\right)}{\left({p}+\mathrm{2}{q}\right)\left({p}+\mathrm{3}{q}\right)}=\mathrm{1} \\ $$

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