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Solve-for-real-numbers-16-x-4-x-1-x-4-x-2-x-1-x-8-x-1-65-

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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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