Question Number 189630 by mathocean1 last updated on 19/Mar/23
$$\mathrm{16}^{{x}} +\mathrm{20}^{{x}} =\mathrm{25}^{{x}} \\ $$$${x}\:\left(\in\:\mathbb{R}\right)\:=? \\ $$
Answered by Rasheed.Sindhi last updated on 19/Mar/23
$$\mathrm{16}^{{x}} +\mathrm{20}^{{x}} =\mathrm{25}^{{x}} \:\:;\:\:\:{x}\:\left(\in\:\mathbb{R}\right)\:=? \\ $$$$\left(\frac{\mathrm{16}}{\mathrm{20}}\right)^{{x}} +\mathrm{1}=\left(\frac{\mathrm{25}}{\mathrm{20}}\right)^{{x}} \\ $$$$\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{{x}} +\mathrm{1}=\left(\frac{\mathrm{5}}{\mathrm{4}}\right)^{{x}} \\ $$$$\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{{x}} −\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{−{x}} +\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:{y}−{y}^{−\mathrm{1}} +\mathrm{1}=\mathrm{0} \\ $$$$\:\:{y}^{\mathrm{2}} +{y}−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:{y}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}}}{\mathrm{2}}\:=\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}\:}{\mathrm{2}} \\ $$$$\:\:\:\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{{x}} =\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}\:}{\mathrm{2}} \\ $$$${x}\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{4}}{\mathrm{5}}\right)=\mathrm{log}_{\mathrm{2}} \left(\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}\:}{\mathrm{2}}\right)=\mathrm{log}_{\mathrm{2}} \left(−\mathrm{1}\pm\sqrt{\mathrm{5}}\:\right)−\mathrm{log}_{\mathrm{2}} \mathrm{2}\:\: \\ $$$${x}=\frac{\mathrm{log}_{\mathrm{2}} \left(−\mathrm{1}\pm\sqrt{\mathrm{5}}\:\right)−\mathrm{1}}{\mathrm{log}_{\mathrm{2}} \mathrm{4}−\mathrm{log}_{\mathrm{2}} \mathrm{5}\:\:} \\ $$$${x}=\frac{\mathrm{log}_{\mathrm{2}} \left(−\mathrm{1}\pm\sqrt{\mathrm{5}}\:\right)−\mathrm{1}}{\mathrm{2}−\mathrm{log}_{\mathrm{2}} \mathrm{5}\:\:} \\ $$$$\because\:\mathrm{log}_{\mathrm{2}} \left(−\mathrm{1}−\sqrt{\mathrm{5}}\:\right)\notin\mathbb{R} \\ $$$$\therefore\:{x}=\frac{\mathrm{log}_{\mathrm{2}} \left(−\mathrm{1}+\sqrt{\mathrm{5}}\:\right)−\mathrm{1}}{\mathrm{2}−\mathrm{log}_{\mathrm{2}} \mathrm{5}\:\:} \\ $$
Commented by Rasheed.Sindhi last updated on 19/Mar/23
$$\mathcal{R}{ight}\:{sir}!\:{Corrected}\:{now}. \\ $$
Commented by JDamian last updated on 19/Mar/23
$$\mathrm{but}\:\mathrm{log}_{\mathrm{2}} \left(−\mathrm{1}−\sqrt{\mathrm{5}}\right)\:\notin\:\mathbb{R} \\ $$
Commented by mathocean1 last updated on 19/Mar/23
$${thanks} \\ $$
Answered by Rasheed.Sindhi last updated on 20/Mar/23
$$\mathbb{A}\boldsymbol{\mathrm{n}}\mathbb{O}\boldsymbol{\mathrm{ther}}\:\mathbb{W}\boldsymbol{\mathrm{ay}} \\ $$$$\mathrm{16}^{{x}} +\mathrm{20}^{{x}} =\mathrm{25}^{{x}} \\ $$$$\left(\frac{\mathrm{16}}{\mathrm{25}}\right)^{{x}} +\left(\frac{\mathrm{20}}{\mathrm{25}}\right)^{{x}} =\mathrm{1} \\ $$$$\left(\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{5}^{\mathrm{2}} }\right)^{{x}} +\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{{x}} =\mathrm{1} \\ $$$$\left\{\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{{x}} \right\}^{\mathrm{2}} +\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{{x}} =\mathrm{1} \\ $$$$\:\:\:\:\:{y}^{\mathrm{2}} +{y}−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:{y}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}}}{\mathrm{2}}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\:\:\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{{x}} =\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{4}}{\mathrm{5}}\right)=\mathrm{log}_{\mathrm{2}} \left(\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\:\: \\ $$$${x}=\frac{\mathrm{log}_{\mathrm{2}} \left(−\mathrm{1}\pm\sqrt{\mathrm{5}}\right)−\mathrm{log}_{\mathrm{2}} \mathrm{2}\:\:}{\mathrm{log}_{\mathrm{2}} \mathrm{4}−\mathrm{log}_{\mathrm{2}} \mathrm{5}\:\:} \\ $$$$\because\:\mathrm{log}_{\mathrm{2}} \left(−\mathrm{1}−\sqrt{\mathrm{5}}\right)\notin\mathbb{R} \\ $$$$\:\therefore\:{x}=\frac{\mathrm{log}_{\mathrm{2}} \left(−\mathrm{1}+\sqrt{\mathrm{5}}\right)−\mathrm{1}\:\:}{\mathrm{2}−\mathrm{log}_{\mathrm{2}} \mathrm{5}\:\:} \\ $$