Question Number 221863 by fantastic last updated on 11/Jun/25
$${If}\:{a}\:{and}\:{b}\:{are}\:{whole}\:{numbers}\:{such}\:{a}^{{b}} =\mathrm{121} \\ $$$${then}\:{find}\:{the}\:{value}\:{of}\:\left({a}−\mathrm{1}\right)^{{b}+\mathrm{1}} \\ $$
Commented by Tawa11 last updated on 11/Jun/25
$$\mathrm{11}^{\mathrm{2}} \:\:=\:\:\mathrm{121} \\ $$$$\mathrm{a}\:=\:\mathrm{11},\:\mathrm{b}\:=\:\mathrm{2} \\ $$$$\therefore\:\:\:\left(\mathrm{11}\:−\:\mathrm{1}\right)^{\mathrm{2}\:+\:\mathrm{1}} \:\:=\:\:\mathrm{10}^{\mathrm{3}} \:\:=\:\:\mathrm{1000} \\ $$
Answered by Rasheed.Sindhi last updated on 11/Jun/25
$${a}^{{b}} =\mathrm{121}^{\mathrm{1}} =\mathrm{11}^{\mathrm{2}} \\ $$$${a}=\mathrm{121}\Rightarrow{b}=\mathrm{1} \\ $$$$\:\:\:\:\:\left({a}−\mathrm{1}\right)^{{b}+\mathrm{1}} =\left(\mathrm{121}−\mathrm{1}\right)^{\mathrm{1}+\mathrm{1}} =\mathrm{14400} \\ $$$${a}=\mathrm{11}\Rightarrow{b}=\mathrm{2} \\ $$$$\:\:\:\:\:\left({a}−\mathrm{1}\right)^{{b}+\mathrm{1}} =\left(\mathrm{11}−\mathrm{1}\right)^{\mathrm{2}+\mathrm{1}} =\mathrm{10}^{\mathrm{3}} =\mathrm{1000} \\ $$$$ \\ $$