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Question Number 92283 by jagoll last updated on 06/May/20

9^x +3^x  = 25^x −5^x    find (5^x /(3^x +1)) ?

$$\mathrm{9}^{\mathrm{x}} +\mathrm{3}^{\mathrm{x}} \:=\:\mathrm{25}^{\mathrm{x}} −\mathrm{5}^{\mathrm{x}} \: \\ $$$$\mathrm{find}\:\frac{\mathrm{5}^{\mathrm{x}} }{\mathrm{3}^{\mathrm{x}} +\mathrm{1}}\:? \\ $$

Commented by john santu last updated on 06/May/20

(3^x )^2 +(3^x )−(5^x )^2 +(5^x )=0  let 3^x  = u ; 5^x =v  (u−v)(u+v)+(u+v)=0  (u+v)(u−v+1)=  ⇒ u = v−1 , v = u+1  5^(x )  = 3^x +1 ⇒ (5^x /(3^x +1)) = 1

$$\left(\mathrm{3}^{{x}} \right)^{\mathrm{2}} +\left(\mathrm{3}^{{x}} \right)−\left(\mathrm{5}^{{x}} \right)^{\mathrm{2}} +\left(\mathrm{5}^{{x}} \right)=\mathrm{0} \\ $$$${let}\:\mathrm{3}^{{x}} \:=\:{u}\:;\:\mathrm{5}^{{x}} ={v} \\ $$$$\left({u}−{v}\right)\left({u}+{v}\right)+\left({u}+{v}\right)=\mathrm{0} \\ $$$$\left({u}+{v}\right)\left({u}−{v}+\mathrm{1}\right)= \\ $$$$\Rightarrow\:{u}\:=\:{v}−\mathrm{1}\:,\:{v}\:=\:{u}+\mathrm{1} \\ $$$$\mathrm{5}^{{x}\:} \:=\:\mathrm{3}^{{x}} +\mathrm{1}\:\Rightarrow\:\frac{\mathrm{5}^{{x}} }{\mathrm{3}^{{x}} +\mathrm{1}}\:=\:\mathrm{1} \\ $$$$ \\ $$

Commented by jagoll last updated on 06/May/20

cool man ������

Commented by $@ty@m123 last updated on 06/May/20

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