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Question Number 96242 by joki last updated on 30/May/20

find x if 4^x +6^x =9^x findx?

$$\mathrm{find}\:\mathrm{x}\:\mathrm{if}\:\mathrm{4}^{\mathrm{x}} +\mathrm{6}^{\mathrm{x}} =\mathrm{9}^{\mathrm{x}} \:\:\:\:\mathrm{findx}? \\ $$

Answered by john santu last updated on 30/May/20

((4/9))^x +((6/9))^x =1 ((2/3))^(2x) +((2/3))^x =1 t^2 +t−1 = 0 , t = ((2/3))^x t = ((−1+(√5))/2) ⇒ x = log _(((2/3))) ((((√5)−1)/2))

$$\left(\frac{\mathrm{4}}{\mathrm{9}}\right)^{{x}} +\left(\frac{\mathrm{6}}{\mathrm{9}}\right)^{{x}} =\mathrm{1} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}{x}} +\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\mathrm{1} \\ $$$${t}^{\mathrm{2}} +{t}−\mathrm{1}\:=\:\mathrm{0}\:,\:\mathrm{t}\:=\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} \\ $$$${t}\:=\:\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow\:{x}\:=\:\mathrm{log}\:_{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)} \left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right) \\ $$

Answered by 1549442205 last updated on 31/May/20

Dividing two sides of the equation by 6^x we get ((2/3))^x +1=((3/2))^x .Putting ((2/3))^x =t (t>0),we obtain: t−(1/t)+1=0⇔t^2 +t−1=0 t=((−1+(√5))/2).From that ((2/3))^x =(((√5)−1)/2) xln((2/3))=ln(((√5)−1)/2)⇒x=((ln(((√5)−1)/2))/(ln(2/3)))≈1.1868

$$\mathrm{Dividing}\:\mathrm{two}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{by} \\ $$$$\mathrm{6}^{\mathrm{x}} \:\mathrm{we}\:\mathrm{get}\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{x}} +\mathrm{1}=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{x}} .\mathrm{Putting}\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{x}} \\ $$$$=\mathrm{t}\:\left(\mathrm{t}>\mathrm{0}\right),\mathrm{we}\:\mathrm{obtain}:\:\mathrm{t}−\frac{\mathrm{1}}{\mathrm{t}}+\mathrm{1}=\mathrm{0}\Leftrightarrow\mathrm{t}^{\mathrm{2}} +\mathrm{t}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{t}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}.\mathrm{From}\:\mathrm{that}\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{x}} =\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{xln}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)=\mathrm{ln}\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{x}=\frac{\mathrm{ln}\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}}{\mathrm{ln}\frac{\mathrm{2}}{\mathrm{3}}}\approx\mathrm{1}.\mathrm{1868} \\ $$$$ \\ $$

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