Question-74168

Question Number 74168 by mr W last updated on 19/Nov/19

Commented by mr W last updated on 19/Nov/19

$${solve}\:{for}\:{x}. \\ $$

Answered by Rio Michael last updated on 19/Nov/19

dividing by 4^x , 1 + ((6/4))^x = ((9/4))^x 1 + ((3/2))^x = ((3/2))^(2x) let ((3/2))^x = a ⇒ 1 + a = a^2 a^2 −a−1=0 a = ((1±(√5))/2) or ((3/2))^x = ((1±(√5))/2) x (log(3/2)) = log[((1±(√5))/2)] x = ((log[((1 ±(√5))/2)])/(log(3/2))) x ≈ 1.2

$${dividing}\:{by}\:\mathrm{4}^{{x}} , \\ $$$$\:\:\:\:\:\:\mathrm{1}\:+\:\left(\frac{\mathrm{6}}{\mathrm{4}}\right)^{{x}} \:=\:\left(\frac{\mathrm{9}}{\mathrm{4}}\right)^{{x}} \\ $$$$\:\:\:\:\:\:\mathrm{1}\:+\:\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}} \:=\:\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}{x}} \\ $$$${let}\:\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}} \:=\:{a} \\ $$$$\Rightarrow\:\mathrm{1}\:+\:{a}\:=\:{a}^{\mathrm{2}} \\ $$$$\:\:{a}^{\mathrm{2}} −{a}−\mathrm{1}=\mathrm{0} \\ $$$$\:\:{a}\:=\:\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${or}\:\:\:\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}} \:=\:\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\:{x}\:\left({log}\frac{\mathrm{3}}{\mathrm{2}}\right)\:=\:{log}\left[\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\right] \\ $$$${x}\:=\:\frac{{log}\left[\frac{\mathrm{1}\:\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\right]}{{log}\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${x}\:\approx\:\mathrm{1}.\mathrm{2} \\ $$

Answered by Maclaurin Stickker last updated on 19/Nov/19

I just saw this question a few years ago in a pdf. 2^(2x) +2^x .3^x −3^(2x) =0 divide by 3^(2x) ((2/3))^(2x) +((2/3))^x −1=0 ((2/3))^x =y y^2 +y−1=0 y=((−1∓(√5))/2)⇒y_1 =((−1+(√5))/2) and y_2 =((−1−(√5))/2) ((2/3))^x =((−1−(√5))/2)⇒x∉R ((2/3))^x =((−1+(√5))/2)⇒x=log_(2/3) ((−1+(√5))/2) S={log_(2/3) (((−1+(√5))/2))}

$${I}\:{just}\:{saw}\:{this}\:{question}\:{a}\:{few}\:{years} \\ $$$${ago}\:{in}\:{a}\:{pdf}. \\ $$$$ \\ $$$$\mathrm{2}^{\mathrm{2}{x}} +\mathrm{2}^{{x}} .\mathrm{3}^{{x}} −\mathrm{3}^{\mathrm{2}{x}} =\mathrm{0} \\ $$$${divide}\:{by}\:\mathrm{3}^{\mathrm{2}{x}} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}{x}} +\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} −\mathrm{1}=\mathrm{0} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} ={y} \\ $$$${y}^{\mathrm{2}} +{y}−\mathrm{1}=\mathrm{0} \\ $$$${y}=\frac{−\mathrm{1}\mp\sqrt{\mathrm{5}}}{\mathrm{2}}\Rightarrow{y}_{\mathrm{1}} =\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:{and}\:{y}_{\mathrm{2}} =\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$ \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\Rightarrow{x}\notin\mathbb{R} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\Rightarrow{x}={log}_{\frac{\mathrm{2}}{\mathrm{3}}} \frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${S}=\left\{{log}_{\frac{\mathrm{2}}{\mathrm{3}}} \left(\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\right\} \\ $$

Commented by Maclaurin Stickker last updated on 19/Nov/19

$${Is}\:{there}\:{another}\:{solution}\:{to}\:{the}\:{problem}? \\ $$

Commented by malwaan last updated on 20/Nov/19

$${yes} \\ $$$${complex}\:{solution} \\ $$

Answered by MJS last updated on 20/Nov/19

4^x +6^x −9^x =0 t=((3/2))^x ∧t>0 ⇔ x=((ln t)/(ln (3/2)))=((ln t)/(ln 3 −ln 2)) (−t^2 +t+1)t^((2ln 2)/(ln 3 −ln 2)) =0 ∧ t>0 ⇒ t=((1+(√5))/2) ⇒ x=((ln (1+(√5)) −ln 2)/(ln 3 −ln 2)) but if we allow x∈C ⇒ t∈C ⇒ t_1 =((1−(√5))/2) ⇒ x_1 =−x_3 +(π/(ln 3 −ln 2))i t_2 =0 ⇒ x_2 =−∞ t_3 =((1+(√5))/2) ⇒ x_3 =((ln (1+(√5)) −ln 2)/(ln 3 −ln 2))

$$\mathrm{4}^{{x}} +\mathrm{6}^{{x}} −\mathrm{9}^{{x}} =\mathrm{0} \\ $$$${t}=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}} \wedge{t}>\mathrm{0}\:\Leftrightarrow\:{x}=\frac{\mathrm{ln}\:{t}}{\mathrm{ln}\:\frac{\mathrm{3}}{\mathrm{2}}}=\frac{\mathrm{ln}\:{t}}{\mathrm{ln}\:\mathrm{3}\:−\mathrm{ln}\:\mathrm{2}} \\ $$$$\left(−{t}^{\mathrm{2}} +{t}+\mathrm{1}\right){t}^{\frac{\mathrm{2ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}\:−\mathrm{ln}\:\mathrm{2}}} =\mathrm{0}\:\wedge\:{t}>\mathrm{0} \\ $$$$\Rightarrow\:{t}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{x}=\frac{\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)\:−\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}\:−\mathrm{ln}\:\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{but}\:\mathrm{if}\:\mathrm{we}\:\mathrm{allow}\:{x}\in\mathbb{C} \\ $$$$\Rightarrow\:{t}\in\mathbb{C} \\ $$$$\Rightarrow \\ $$$${t}_{\mathrm{1}} =\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow\:{x}_{\mathrm{1}} =−{x}_{\mathrm{3}} +\frac{\pi}{\mathrm{ln}\:\mathrm{3}\:−\mathrm{ln}\:\mathrm{2}}\mathrm{i} \\ $$$${t}_{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:{x}_{\mathrm{2}} =−\infty \\ $$$${t}_{\mathrm{3}} =\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow\:{x}_{\mathrm{3}} =\frac{\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)\:−\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}\:−\mathrm{ln}\:\mathrm{2}} \\ $$

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