Menu Close

If-a-x-b-y-a-x-b-y-1-then-a-x-b-y-




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

Leave a Reply

Your email address will not be published. Required fields are marked *