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Question Number 189630 by mathocean1 last updated on 19/Mar/23

16^x +20^x =25^x x (∈ R) =?

$$\mathrm{16}^{{x}} +\mathrm{20}^{{x}} =\mathrm{25}^{{x}} \\ $$$${x}\:\left(\in\:\mathbb{R}\right)\:=? \\ $$

Answered by Rasheed.Sindhi last updated on 19/Mar/23

16^x +20^x =25^x ; x (∈ R) =? (((16)/(20)))^x +1=(((25)/(20)))^x ((4/5))^x +1=((5/4))^x ((4/5))^x −((4/5))^(−x) +1=0 y−y^(−1) +1=0 y^2 +y−1=0 y=((−1±(√(1+4)))/2) =((−1±(√5) )/2) ((4/5))^x =((−1±(√5) )/2) xlog_2 ((4/5))=log_2 (((−1±(√5) )/2))=log_2 (−1±(√5) )−log_2 2 x=((log_2 (−1±(√5) )−1)/(log_2 4−log_2 5 )) x=((log_2 (−1±(√5) )−1)/(2−log_2 5 )) ∵ log_2 (−1−(√5) )∉R ∴ x=((log_2 (−1+(√5) )−1)/(2−log_2 5 ))

$$\mathrm{16}^{{x}} +\mathrm{20}^{{x}} =\mathrm{25}^{{x}} \:\:;\:\:\:{x}\:\left(\in\:\mathbb{R}\right)\:=? \\ $$$$\left(\frac{\mathrm{16}}{\mathrm{20}}\right)^{{x}} +\mathrm{1}=\left(\frac{\mathrm{25}}{\mathrm{20}}\right)^{{x}} \\ $$$$\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{{x}} +\mathrm{1}=\left(\frac{\mathrm{5}}{\mathrm{4}}\right)^{{x}} \\ $$$$\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{{x}} −\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{−{x}} +\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:{y}−{y}^{−\mathrm{1}} +\mathrm{1}=\mathrm{0} \\ $$$$\:\:{y}^{\mathrm{2}} +{y}−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:{y}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}}}{\mathrm{2}}\:=\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}\:}{\mathrm{2}} \\ $$$$\:\:\:\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{{x}} =\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}\:}{\mathrm{2}} \\ $$$${x}\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{4}}{\mathrm{5}}\right)=\mathrm{log}_{\mathrm{2}} \left(\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}\:}{\mathrm{2}}\right)=\mathrm{log}_{\mathrm{2}} \left(−\mathrm{1}\pm\sqrt{\mathrm{5}}\:\right)−\mathrm{log}_{\mathrm{2}} \mathrm{2}\:\: \\ $$$${x}=\frac{\mathrm{log}_{\mathrm{2}} \left(−\mathrm{1}\pm\sqrt{\mathrm{5}}\:\right)−\mathrm{1}}{\mathrm{log}_{\mathrm{2}} \mathrm{4}−\mathrm{log}_{\mathrm{2}} \mathrm{5}\:\:} \\ $$$${x}=\frac{\mathrm{log}_{\mathrm{2}} \left(−\mathrm{1}\pm\sqrt{\mathrm{5}}\:\right)−\mathrm{1}}{\mathrm{2}−\mathrm{log}_{\mathrm{2}} \mathrm{5}\:\:} \\ $$$$\because\:\mathrm{log}_{\mathrm{2}} \left(−\mathrm{1}−\sqrt{\mathrm{5}}\:\right)\notin\mathbb{R} \\ $$$$\therefore\:{x}=\frac{\mathrm{log}_{\mathrm{2}} \left(−\mathrm{1}+\sqrt{\mathrm{5}}\:\right)−\mathrm{1}}{\mathrm{2}−\mathrm{log}_{\mathrm{2}} \mathrm{5}\:\:} \\ $$

Commented by Rasheed.Sindhi last updated on 19/Mar/23

Right sir! Corrected now.

$$\mathcal{R}{ight}\:{sir}!\:{Corrected}\:{now}. \\ $$

Commented by JDamian last updated on 19/Mar/23

but log_2 (−1−(√5)) ∉ R

$$\mathrm{but}\:\mathrm{log}_{\mathrm{2}} \left(−\mathrm{1}−\sqrt{\mathrm{5}}\right)\:\notin\:\mathbb{R} \\ $$

Commented by mathocean1 last updated on 19/Mar/23

thanks

$${thanks} \\ $$

Answered by Rasheed.Sindhi last updated on 20/Mar/23

AnOther Way 16^x +20^x =25^x (((16)/(25)))^x +(((20)/(25)))^x =1 ((4^2 /5^2 ))^x +((4/5))^x =1 {((4/5))^x }^2 +((4/5))^x =1 y^2 +y−1=0 y=((−1±(√(1+4)))/2)=((−1±(√5))/2) ((4/5))^x =((−1±(√5))/2) xlog_2 ((4/5))=log_2 (((−1±(√5))/2)) x=((log_2 (−1±(√5))−log_2 2 )/(log_2 4−log_2 5 )) ∵ log_2 (−1−(√5))∉R ∴ x=((log_2 (−1+(√5))−1 )/(2−log_2 5 ))

$$\mathbb{A}\boldsymbol{\mathrm{n}}\mathbb{O}\boldsymbol{\mathrm{ther}}\:\mathbb{W}\boldsymbol{\mathrm{ay}} \\ $$$$\mathrm{16}^{{x}} +\mathrm{20}^{{x}} =\mathrm{25}^{{x}} \\ $$$$\left(\frac{\mathrm{16}}{\mathrm{25}}\right)^{{x}} +\left(\frac{\mathrm{20}}{\mathrm{25}}\right)^{{x}} =\mathrm{1} \\ $$$$\left(\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{5}^{\mathrm{2}} }\right)^{{x}} +\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{{x}} =\mathrm{1} \\ $$$$\left\{\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{{x}} \right\}^{\mathrm{2}} +\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{{x}} =\mathrm{1} \\ $$$$\:\:\:\:\:{y}^{\mathrm{2}} +{y}−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:{y}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}}}{\mathrm{2}}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\:\:\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{{x}} =\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{4}}{\mathrm{5}}\right)=\mathrm{log}_{\mathrm{2}} \left(\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\:\: \\ $$$${x}=\frac{\mathrm{log}_{\mathrm{2}} \left(−\mathrm{1}\pm\sqrt{\mathrm{5}}\right)−\mathrm{log}_{\mathrm{2}} \mathrm{2}\:\:}{\mathrm{log}_{\mathrm{2}} \mathrm{4}−\mathrm{log}_{\mathrm{2}} \mathrm{5}\:\:} \\ $$$$\because\:\mathrm{log}_{\mathrm{2}} \left(−\mathrm{1}−\sqrt{\mathrm{5}}\right)\notin\mathbb{R} \\ $$$$\:\therefore\:{x}=\frac{\mathrm{log}_{\mathrm{2}} \left(−\mathrm{1}+\sqrt{\mathrm{5}}\right)−\mathrm{1}\:\:}{\mathrm{2}−\mathrm{log}_{\mathrm{2}} \mathrm{5}\:\:} \\ $$

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