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Question Number 164124 by HongKing last updated on 14/Jan/22

Solve for real numbers:  ((16^x  + 4^x  + 1^x )/(4^x  + 2^x  + 1^x )) = ((8^x  + 1)/(65))

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\frac{\mathrm{16}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{4}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{1}^{\boldsymbol{\mathrm{x}}} }{\mathrm{4}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{1}^{\boldsymbol{\mathrm{x}}} }\:=\:\frac{\mathrm{8}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{1}}{\mathrm{65}} \\ $$

Answered by MJS_new last updated on 14/Jan/22

((t^4 +t^2 +1)/(t^2 +t+1))=((t^3 +1)/(65))  (((t^2 +t+1)(t^2 −t+1))/(t^2 +t+1))=(((t+1)(t^2 −t+1))/(65))  1=((t+1)/(65))  t=64  x=8

$$\frac{{t}^{\mathrm{4}} +{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}=\frac{{t}^{\mathrm{3}} +\mathrm{1}}{\mathrm{65}} \\ $$$$\frac{\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}=\frac{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)}{\mathrm{65}} \\ $$$$\mathrm{1}=\frac{{t}+\mathrm{1}}{\mathrm{65}} \\ $$$${t}=\mathrm{64} \\ $$$${x}=\mathrm{8} \\ $$

Commented by HongKing last updated on 15/Jan/22

thank you so much my dear Sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir} \\ $$

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