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Solve-for-x-in-the-equation-3-x-4-x-5-x-6-x-




Question Number 10360 by Tawakalitu ayo mi last updated on 05/Feb/17
Solve for x in the equation.  3^x  + 4^x  + 5^x  = 6^x
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}\:\mathrm{in}\:\mathrm{the}\:\mathrm{equation}. \\ $$$$\mathrm{3}^{\mathrm{x}} \:+\:\mathrm{4}^{\mathrm{x}} \:+\:\mathrm{5}^{\mathrm{x}} \:=\:\mathrm{6}^{\mathrm{x}} \\ $$
Answered by mrW1 last updated on 05/Feb/17
when x=2,  3^2 +4^2 +5^2 =9+16+25=50  6^2 =36  i.e. 3^2 +4^2 +5^2 >6^2     when x=4,  3^4 +4^4 +5^4 =81+256+625=962  6^4 =1296  i.e. 3^4 +4^4 +5^4 <6^4     ⇒the solution is between 2 and 4.    through try and error we get the  solution x=3:  3^3 +4^3 +5^3 =27+64+125=216  6^3 =216
$${when}\:{x}=\mathrm{2}, \\ $$$$\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} =\mathrm{9}+\mathrm{16}+\mathrm{25}=\mathrm{50} \\ $$$$\mathrm{6}^{\mathrm{2}} =\mathrm{36} \\ $$$${i}.{e}.\:\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} >\mathrm{6}^{\mathrm{2}} \\ $$$$ \\ $$$${when}\:{x}=\mathrm{4}, \\ $$$$\mathrm{3}^{\mathrm{4}} +\mathrm{4}^{\mathrm{4}} +\mathrm{5}^{\mathrm{4}} =\mathrm{81}+\mathrm{256}+\mathrm{625}=\mathrm{962} \\ $$$$\mathrm{6}^{\mathrm{4}} =\mathrm{1296} \\ $$$${i}.{e}.\:\mathrm{3}^{\mathrm{4}} +\mathrm{4}^{\mathrm{4}} +\mathrm{5}^{\mathrm{4}} <\mathrm{6}^{\mathrm{4}} \\ $$$$ \\ $$$$\Rightarrow{the}\:{solution}\:{is}\:{between}\:\mathrm{2}\:{and}\:\mathrm{4}. \\ $$$$ \\ $$$${through}\:{try}\:{and}\:{error}\:{we}\:{get}\:{the} \\ $$$${solution}\:{x}=\mathrm{3}: \\ $$$$\mathrm{3}^{\mathrm{3}} +\mathrm{4}^{\mathrm{3}} +\mathrm{5}^{\mathrm{3}} =\mathrm{27}+\mathrm{64}+\mathrm{125}=\mathrm{216} \\ $$$$\mathrm{6}^{\mathrm{3}} =\mathrm{216} \\ $$
Commented by mrW1 last updated on 05/Feb/17
I′m not sure if there is an analytical  solution to this question.
$${I}'{m}\:{not}\:{sure}\:{if}\:{there}\:{is}\:{an}\:{analytical} \\ $$$${solution}\:{to}\:{this}\:{question}. \\ $$
Commented by Tawakalitu ayo mi last updated on 05/Feb/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by arge last updated on 05/Feb/17
3^x +4^x =6^x −5^x     3^x =6^x   4^x = −5^x     ±4^x = −5^x   −4^x = −5^x     artificio,    3^x −4^x =6^x −5^x   x=0∵∵∵∵Rta
$$\mathrm{3}^{{x}} +\mathrm{4}^{{x}} =\mathrm{6}^{{x}} −\mathrm{5}^{{x}} \\ $$$$ \\ $$$$\mathrm{3}^{{x}} =\mathrm{6}^{{x}} \\ $$$$\mathrm{4}^{{x}} =\:−\mathrm{5}^{{x}} \\ $$$$ \\ $$$$\pm\mathrm{4}^{{x}} =\:−\mathrm{5}^{{x}} \\ $$$$−\mathrm{4}^{{x}} =\:−\mathrm{5}^{{x}} \\ $$$$ \\ $$$${artificio}, \\ $$$$ \\ $$$$\mathrm{3}^{{x}} −\mathrm{4}^{{x}} =\mathrm{6}^{{x}} −\mathrm{5}^{{x}} \\ $$$${x}=\mathrm{0}\because\because\because\because{Rta} \\ $$
Commented by FilupSmith last updated on 06/Feb/17
if x=0, 3=1
$$\mathrm{if}\:{x}=\mathrm{0},\:\mathrm{3}=\mathrm{1} \\ $$
Answered by chux last updated on 06/Feb/17
        i think it can be solved by linear approximation    (1+2)^x +(1+3)^x +(1+4)^x =(1+5)^x
$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\mathrm{i}\:\mathrm{think}\:\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{by}\:\mathrm{linear}\:\mathrm{approximation} \\ $$$$ \\ $$$$\left(\mathrm{1}+\mathrm{2}\right)^{\mathrm{x}} +\left(\mathrm{1}+\mathrm{3}\right)^{\mathrm{x}} +\left(\mathrm{1}+\mathrm{4}\right)^{\mathrm{x}} =\left(\mathrm{1}+\mathrm{5}\right)^{\mathrm{x}} \\ $$

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