16^x +20^x =25^x x (∈ R) =?


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

$16^{x} + 20^{x} = 25^{x}$ $x (\in R) = ?$
	Answered by Rasheed.Sindhi last updated on 19/Mar/23

	$16^{x} + 20^{x} = 25^{x}; x (\in R) = ?$ ${(\frac{16}{20})}^{x} + 1 = {(\frac{25}{20})}^{x}$ ${(\frac{4}{5})}^{x} + 1 = {(\frac{5}{4})}^{x}$ ${(\frac{4}{5})}^{x} - {(\frac{4}{5})}^{- x} + 1 = 0$ $y - y^{- 1} + 1 = 0$ $y^{2} + y - 1 = 0$ $y = \frac{- 1 \pm \sqrt{1 + 4}}{2} = \frac{- 1 \pm \sqrt{5}}{2}$ ${(\frac{4}{5})}^{x} = \frac{- 1 \pm \sqrt{5}}{2}$ $x \log_{2} (\frac{4}{5}) = \log_{2} (\frac{- 1 \pm \sqrt{5}}{2}) = \log_{2} (- 1 \pm \sqrt{5}) - \log_{2} 2$ $x = \frac{\log_{2} (- 1 \pm \sqrt{5}) - 1}{\log_{2} 4 - \log_{2} 5}$ $x = \frac{\log_{2} (- 1 \pm \sqrt{5}) - 1}{2 - \log_{2} 5}$ $∵ \log_{2} (- 1 - \sqrt{5}) \notin R$ $∴ x = \frac{\log_{2} (- 1 + \sqrt{5}) - 1}{2 - \log_{2} 5}$
	Commented by Rasheed.Sindhi last updated on 19/Mar/23

	$R i g h t s i r! C o r r e c t e d n o w .$
	Commented by JDamian last updated on 19/Mar/23

	$but \log_{2} (- 1 - \sqrt{5}) \notin R$
	Commented by mathocean1 last updated on 19/Mar/23

	$t h a n k s$
	Answered by Rasheed.Sindhi last updated on 20/Mar/23
	$AnOther Way 16^x +20^x =25^x (((16)/(25)))^x +(((20)/(25)))^x =1 ((4^2 /5^2 ))^x +((4/5))^x =1 {((4/5))^x }^2 +((4/5))^x =1 y^2 +y−1=0 y=((−1±(√(1+4)))/2)=((−1±(√5))/2) ((4/5))^x =((−1±(√5))/2) xlog_2 ((4/5))=log_2 (((−1±(√5))/2)) x=((log_2 (−1±(√5))−log_2 2 )/(log_2 4−log_2 5 )) ∵ log_2 (−1−(√5))∉R ∴ x=((log_2 (−1+(√5))−1 )/(2−log_2 5 ))$
	$A n O ther W ay$ $16^{x} + 20^{x} = 25^{x}$ ${(\frac{16}{25})}^{x} + {(\frac{20}{25})}^{x} = 1$ ${(\frac{4^{2}}{5^{2}})}^{x} + {(\frac{4}{5})}^{x} = 1$ ${{(\frac{4}{5})}^{x}}^{2} + {(\frac{4}{5})}^{x} = 1$ $y^{2} + y - 1 = 0$ $y = \frac{- 1 \pm \sqrt{1 + 4}}{2} = \frac{- 1 \pm \sqrt{5}}{2}$ ${(\frac{4}{5})}^{x} = \frac{- 1 \pm \sqrt{5}}{2}$ $x \log_{2} (\frac{4}{5}) = \log_{2} (\frac{- 1 \pm \sqrt{5}}{2})$ $x = \frac{\log_{2} (- 1 \pm \sqrt{5}) - \log_{2} 2}{\log_{2} 4 - \log_{2} 5}$ $∵ \log_{2} (- 1 - \sqrt{5}) \notin R$ $∴ x = \frac{\log_{2} (- 1 + \sqrt{5}) - 1}{2 - \log_{2} 5}$
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