Question Number 221618 by fantastic last updated on 08/Jun/25
$${solve}\:{for}\:{x} \\ $$$$\mathrm{2}^{{x}} +\mathrm{4}^{{x}} =\mathrm{8}^{{x}} \\ $$
Answered by MathematicalUser2357 last updated on 09/Jun/25
$$\mathrm{log}_{\mathrm{2}} \phi \\ $$
Answered by fantastic last updated on 08/Jun/25
![2^x +4^x =8^x or (2^x /2^x )+(4^x /2^x )=(8^x /2^x ) or 1+2^x =4^x or 1+2^x =2^(2x) let 2^x =u then the equation becomes 1+u=u^2 or u^2 −u−1=0 so u =((−(−1)±(√((−1)^2 −4(1)(−1))))/(2(1))) or u=((1±(√5))/2) but 2^x =u so 2^x =(((1±(√5))/2)) or x=log _2 (((1+(√5))/2))[as ((1−(√5))/2)<0 and the argument must be>0] so x=log _2 (((1+(√5))/2))](https://www.tinkutara.com/question/Q221621.png)
$$\mathrm{2}^{{x}} +\mathrm{4}^{{x}} =\mathrm{8}^{{x}} \\ $$$${or}\:\frac{\mathrm{2}^{{x}} }{\mathrm{2}^{{x}} }+\frac{\mathrm{4}^{{x}} }{\mathrm{2}^{{x}} }=\frac{\mathrm{8}^{{x}} }{\mathrm{2}^{{x}} } \\ $$$${or}\:\mathrm{1}+\mathrm{2}^{{x}} =\mathrm{4}^{{x}} \\ $$$${or}\:\mathrm{1}+\mathrm{2}^{{x}} =\mathrm{2}^{\mathrm{2}{x}} \\ $$$${let}\:\mathrm{2}^{{x}} ={u}\:{then}\:{the}\:{equation}\:{becomes} \\ $$$$\mathrm{1}+{u}={u}^{\mathrm{2}} \\ $$$${or}\:{u}^{\mathrm{2}} −{u}−\mathrm{1}=\mathrm{0} \\ $$$${so}\:{u}\:=\frac{−\left(−\mathrm{1}\right)\pm\sqrt{\left(−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(−\mathrm{1}\right)}}{\mathrm{2}\left(\mathrm{1}\right)} \\ $$$${or}\:{u}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${but}\:\mathrm{2}^{{x}} ={u} \\ $$$${so}\:\mathrm{2}^{{x}} =\left(\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$$${or}\:{x}=\mathrm{log}\underset{\mathrm{2}} {\:}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left[{as}\:\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}<\mathrm{0}\:{and}\:{the}\:{argument}\:{must}\:{be}>\mathrm{0}\right] \\ $$$${so}\:\underline{\boldsymbol{{x}}=\boldsymbol{\mathrm{log}}\underset{\mathrm{2}} {\:}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)} \\ $$
Answered by Frix last updated on 08/Jun/25
$$\mathrm{8}^{{x}} −\mathrm{4}^{{x}} −\mathrm{2}^{{x}} =\mathrm{0} \\ $$$$\mathrm{2}^{{x}} ={t}>\mathrm{0} \\ $$$${t}^{\mathrm{3}} −{t}^{\mathrm{2}} −{t}=\mathrm{0} \\ $$$${t}\left({t}^{\mathrm{2}} −{t}−\mathrm{1}\right)=\mathrm{0} \\ $$$${t}>\mathrm{0}\:\Rightarrow\:{t}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}=\frac{\mathrm{ln}\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}}{\mathrm{ln}\:\mathrm{2}}=\mathrm{log}_{\mathrm{2}} \:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$
Commented by fantastic last updated on 08/Jun/25
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