Question Number 194499 by Mastermind last updated on 08/Jul/23
$$\mathrm{3}^{\mathrm{x}} \:+\:\mathrm{4}^{\mathrm{x}} \:=\:\mathrm{5}^{\mathrm{x}} \\ $$$$\mathrm{find}\:\mathrm{x}\:? \\ $$
Commented by Frix last updated on 08/Jul/23
$$\mathrm{I}\:\mathrm{am}\:\mathrm{a}\:\mathrm{bit}\:\mathrm{blind}\:\mathrm{today}...\:\mathrm{just}\:\mathrm{guessing}: \\ $$$$\mathrm{1}.\mathrm{999999}<{x}<\mathrm{2}.\mathrm{000001} \\ $$
Commented by necx122 last updated on 11/Jul/23
I remember this question in those days and I think people like mr. w, ajfour and 12345 tackled them back then in 2014,2015,2016. I think one of the approaches those times was also linear approximation to the first approximate value. (a + b)^x = a^x + xCo(a)^(x-1)(b)+........
Commented by Frix last updated on 11/Jul/23
$$\mathrm{Sorry}\:\mathrm{but}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{method}\:\mathrm{necessary}\:\mathrm{in} \\ $$$$\mathrm{this}\:\mathrm{special}\:\mathrm{case}.\:\mathrm{It}'\mathrm{s}\:\mathrm{just}\:\mathrm{a}\:\mathrm{Pythagorean} \\ $$$$\mathrm{Triple}\:\mathrm{like}\:\mathrm{5}^{{x}} +\mathrm{12}^{{x}} =\mathrm{13}^{{x}} . \\ $$
Answered by Rasheed.Sindhi last updated on 09/Jul/23
$$\mathrm{3}^{\mathrm{x}} \:+\:\mathrm{4}^{\mathrm{x}} \:=\:\mathrm{5}^{\mathrm{x}} \:\:\:\:\:\mathrm{x}=? \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{x}} +\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{\mathrm{x}} =\mathrm{1} \\ $$$${Let}\:\frac{\mathrm{3}}{\mathrm{5}}=\mathrm{sin}\:\theta \\ $$$$\mathrm{cos}\:\theta=\pm\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta\:}\:=\pm\sqrt{\mathrm{1}−\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\pm\sqrt{\frac{\mathrm{16}}{\mathrm{25}}}\:=\pm\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\frac{\mathrm{4}}{\mathrm{5}}=\mathrm{cos}\:\theta \\ $$$$\left(\mathrm{sin}\:\theta\right)^{\mathrm{x}} +\left(\mathrm{cos}\:\theta\right)^{\mathrm{x}} =\mathrm{1}=\mathrm{sin}^{\mathrm{2}} \theta+\mathrm{cos}^{\mathrm{2}} \theta \\ $$$$\Rightarrow\mathrm{x}=\mathrm{2}\:\: \\ $$$$ \\ $$
Commented by Frix last updated on 11/Jul/23
$$\mathrm{This}\:\mathrm{only}\:\mathrm{works}\:\mathrm{for}\:{a}^{{x}} +{b}^{{x}} ={c}^{{x}} \:\wedge\:{x}=\mathrm{2}.\:\mathrm{Try} \\ $$$$\mathrm{3}^{{x}} +\mathrm{4}^{{x}} =\mathrm{6}^{{x}} \\ $$
Commented by Rasheed.Sindhi last updated on 12/Jul/23
$$\mathrm{Right}\:\mathrm{sir}! \\ $$