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Question Number 146610 by mathdanisur last updated on 14/Jul/21
How many roots does the equation  have?  5^(3x)  + 27∙5^(−3x)  + 9∙5^x  + 27∙5^(−x)  = 64
$${How}\:{many}\:{roots}\:{does}\:{the}\:{equation} \\ $$$${have}? \\ $$$$\mathrm{5}^{\mathrm{3}\boldsymbol{{x}}} \:+\:\mathrm{27}\centerdot\mathrm{5}^{−\mathrm{3}\boldsymbol{{x}}} \:+\:\mathrm{9}\centerdot\mathrm{5}^{\boldsymbol{{x}}} \:+\:\mathrm{27}\centerdot\mathrm{5}^{−\boldsymbol{{x}}} \:=\:\mathrm{64} \\ $$
Answered by iloveisrael last updated on 14/Jul/21
 5^(3x) +27.5^(−3x) +9.5^x +27.5^(−x) = 64   solution: set 5^x =t   t^3 +((27)/t^3 )+9t+((27)/t)=64  t^3 +((3/t))^3 +9(t+(3/t))=64  (t+(3/t))(t^2 +(9/t^2 )−3)+9(t+(3/t))=64  (t+(3/t))[(t+(3/t))^2 −9]+9(t+(3/t))=64  set t+(3/t)=u   ⇒u(u^2 −9)+9u−64=0  ⇒u^3 −64=0  ⇒(u−4)(u^2 −4u+16)_(not real value) =0  ⇒t+(3/t)=4  ⇒t^2 −4t+3=0  t=1 ∧ t=3  ⇒5^x =1⇒x=0  ⇒5^x =3⇒x=log _5 (3)
$$\:\mathrm{5}^{\mathrm{3x}} +\mathrm{27}.\mathrm{5}^{−\mathrm{3x}} +\mathrm{9}.\mathrm{5}^{\mathrm{x}} +\mathrm{27}.\mathrm{5}^{−\mathrm{x}} =\:\mathrm{64} \\ $$$$\:\mathrm{solution}:\:\mathrm{set}\:\mathrm{5}^{\mathrm{x}} =\mathrm{t}\: \\ $$$$\mathrm{t}^{\mathrm{3}} +\frac{\mathrm{27}}{\mathrm{t}^{\mathrm{3}} }+\mathrm{9t}+\frac{\mathrm{27}}{\mathrm{t}}=\mathrm{64} \\ $$$$\mathrm{t}^{\mathrm{3}} +\left(\frac{\mathrm{3}}{\mathrm{t}}\right)^{\mathrm{3}} +\mathrm{9}\left(\mathrm{t}+\frac{\mathrm{3}}{\mathrm{t}}\right)=\mathrm{64} \\ $$$$\left(\mathrm{t}+\frac{\mathrm{3}}{\mathrm{t}}\right)\left(\mathrm{t}^{\mathrm{2}} +\frac{\mathrm{9}}{\mathrm{t}^{\mathrm{2}} }−\mathrm{3}\right)+\mathrm{9}\left(\mathrm{t}+\frac{\mathrm{3}}{\mathrm{t}}\right)=\mathrm{64} \\ $$$$\left(\mathrm{t}+\frac{\mathrm{3}}{\mathrm{t}}\right)\left[\left(\mathrm{t}+\frac{\mathrm{3}}{\mathrm{t}}\right)^{\mathrm{2}} −\mathrm{9}\right]+\mathrm{9}\left(\mathrm{t}+\frac{\mathrm{3}}{\mathrm{t}}\right)=\mathrm{64} \\ $$$$\mathrm{set}\:\mathrm{t}+\frac{\mathrm{3}}{\mathrm{t}}=\mathrm{u}\: \\ $$$$\Rightarrow\mathrm{u}\left(\mathrm{u}^{\mathrm{2}} −\mathrm{9}\right)+\mathrm{9u}−\mathrm{64}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{u}^{\mathrm{3}} −\mathrm{64}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{u}−\mathrm{4}\right)\underset{\mathrm{not}\:\mathrm{real}\:\mathrm{value}} {\underbrace{\left(\mathrm{u}^{\mathrm{2}} −\mathrm{4u}+\mathrm{16}\right)}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{t}+\frac{\mathrm{3}}{\mathrm{t}}=\mathrm{4} \\ $$$$\Rightarrow\mathrm{t}^{\mathrm{2}} −\mathrm{4t}+\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{t}=\mathrm{1}\:\wedge\:\mathrm{t}=\mathrm{3} \\ $$$$\Rightarrow\mathrm{5}^{\mathrm{x}} =\mathrm{1}\Rightarrow\mathrm{x}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{5}^{\mathrm{x}} =\mathrm{3}\Rightarrow\mathrm{x}=\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{3}\right) \\ $$
Commented by mathdanisur last updated on 14/Jul/21
Thankyou Ser cool, roots 2 .?
$${Thankyou}\:{Ser}\:{cool},\:{roots}\:\mathrm{2}\:.? \\ $$

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