Question Number 202116 by Calculusboy last updated on 21/Dec/23
$$\sqrt{\mathrm{3}^{\boldsymbol{{x}}} }\:+\mathrm{1}=\mathrm{2}^{\boldsymbol{{x}}} \:\:\:\boldsymbol{{find}}\:\boldsymbol{{x}} \\ $$
Answered by Frix last updated on 21/Dec/23
$$\mathrm{Obviously}\:{x}=\mathrm{2} \\ $$$$\sqrt{\mathrm{3}^{\mathrm{2}} }+\mathrm{1}=\sqrt{\mathrm{9}}+\mathrm{1}=\mathrm{3}+\mathrm{1}=\mathrm{4}=\mathrm{2}^{\mathrm{2}} \\ $$
Commented by Calculusboy last updated on 21/Dec/23
$$\boldsymbol{{thanks}},\boldsymbol{{but}}\:\boldsymbol{{sir}}\:\boldsymbol{{can}}\:\boldsymbol{{i}}\:\boldsymbol{{get}}\:\boldsymbol{{solution}}\:\boldsymbol{{for}}\:\boldsymbol{{it}} \\ $$
Answered by MATHEMATICSAM last updated on 22/Dec/23
$$\sqrt{\mathrm{3}^{\boldsymbol{{x}}} }\:+\mathrm{1}=\mathrm{2}^{\boldsymbol{{x}}} \:\:\:\boldsymbol{{find}}\:\boldsymbol{{x}} \\ $$$$\mathrm{let}\:{f}\left({x}\right)\:=\:\sqrt{\mathrm{3}^{{x}} }\:+\:\mathrm{1}\:−\:\mathrm{2}^{{x}} \\ $$$${f}\left(\mathrm{2}\right)\:=\:\sqrt{\mathrm{3}^{\mathrm{2}} }\:+\:\mathrm{1}\:−\:\mathrm{2}^{\mathrm{2}} \:=\:\mathrm{3}\:+\:\mathrm{1}\:−\:\mathrm{4}\:=\:\mathrm{0} \\ $$$$\mathrm{As}\:{f}\left(\mathrm{2}\right)\:=\:\mathrm{0}\:\mathrm{so}\:\mathrm{that}\:\mathrm{we}\:\mathrm{can}\:\mathrm{say}\:{x}\:=\:\mathrm{2} \\ $$
Commented by Calculusboy last updated on 23/Dec/23
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$
Commented by Frix last updated on 23/Dec/23
$$\mathrm{This}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{method}. \\ $$$$\mathrm{Try}\:\sqrt{\mathrm{5}^{{x}} }+\mathrm{1}=\mathrm{3}^{{x}} \\ $$$$\mathrm{You}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate}. \\ $$$$\sqrt{\mathrm{3}^{{x}} }+\mathrm{1}=\mathrm{2}^{{x}} \:\mathrm{is}\:\mathrm{constructed}\:\mathrm{to}\:\mathrm{have}\:\mathrm{the} \\ $$$$\mathrm{obvious}\:\mathrm{solution}\:{x}=\mathrm{2}.\:\mathrm{The}\:\mathrm{principle}\:\mathrm{is} \\ $$$${a}+{b}={c}^{{n}} \:\Rightarrow\:\sqrt[{{n}}]{{a}^{{n}} }+{b}={c}^{{n}} \\ $$