Question Number 208519 by Tawa11 last updated on 17/Jun/24
$$\mathrm{If}\:\:\:\:\:\mathrm{a}^{\mathrm{x}} \:\:=\:\:\mathrm{b}^{\mathrm{y}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{a}}{\mathrm{x}}\:\:+\:\:\frac{\mathrm{b}}{\mathrm{y}}\:\:=\:\:\mathrm{1} \\ $$$$\mathrm{then},\:\:\:\:\mathrm{a}^{\mathrm{x}} \:\:+\:\:\mathrm{b}^{\mathrm{y}} \:\:=\:\:? \\ $$
Answered by mr W last updated on 17/Jun/24
$${a}^{{x}} ={b}^{{y}} ={k},\:{say} \\ $$$${x}=\frac{\mathrm{log}\:{k}}{\mathrm{log}\:{a}} \\ $$$${y}=\frac{\mathrm{log}\:{k}}{\mathrm{log}\:{b}} \\ $$$$\frac{{a}\:\mathrm{log}\:{a}}{\mathrm{log}\:{k}}+\frac{{b}\:\mathrm{log}\:{b}}{\mathrm{log}\:{k}}=\mathrm{1} \\ $$$${a}\:\mathrm{log}\:{a}+{b}\:\mathrm{log}\:{b}=\mathrm{log}\:{a}^{{a}} {b}^{{b}} =\mathrm{log}\:{k} \\ $$$$\Rightarrow{k}={a}^{{a}} {b}^{{b}} \\ $$$${a}^{{x}} +{b}^{{y}} =\mathrm{2}{k}=\mathrm{2}{a}^{{a}} {b}^{{b}} \:\checkmark \\ $$
Commented by Tawa11 last updated on 17/Jun/24
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$