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16-x-20-x-25-x-x-R-




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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