Algebra Questions

Question Number 105686 by bramlex last updated on 31/Jul/20

$${solve}\:\mathrm{7}^{{x}} \:+\:\mathrm{24}^{{x}} \:=\:\mathrm{25}^{{x}} \: \\$$

Answered by Rasheed.Sindhi last updated on 31/Jul/20

$$\left(\frac{\mathrm{7}}{\mathrm{25}}\right)^{{x}} +\left(\frac{\mathrm{24}}{\mathrm{25}}\right)^{{x}} =\mathrm{1} \\$$$${Let}\:\frac{\mathrm{7}}{\mathrm{25}}=\mathrm{sin}\:\theta, \\$$$$\:\:\:\mathrm{cos}\:\theta=\sqrt{\mathrm{1}−\mathrm{si}{n}^{\mathrm{2}} \theta}=\sqrt{\mathrm{1}−\left(\frac{\mathrm{7}}{\mathrm{25}}\right)^{\mathrm{2}} } \\$$$$\:\:\:\:\:\:\:=\sqrt{\mathrm{1}−\frac{\mathrm{49}}{\mathrm{625}}}\:=\sqrt{\frac{\mathrm{625}−\mathrm{49}}{\mathrm{625}}}=\frac{\mathrm{24}}{\mathrm{25}} \\$$$$\therefore\:\left(\frac{\mathrm{7}}{\mathrm{25}}\right)^{{x}} +\left(\frac{\mathrm{24}}{\mathrm{25}}\right)^{{x}} =\mathrm{1}\Rightarrow \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{sin}\:\theta\right)^{{x}} +\left(\mathrm{cos}\:\theta\right)^{{x}} =\mathrm{1} \\$$$${Comparing}\:{with}\:{the}\:{identity}: \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{sin}\:\theta\right)^{\mathrm{2}} +\left(\mathrm{cos}\:\theta\right)^{\mathrm{2}} =\mathrm{1} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}=\mathrm{2} \\$$

Commented by bramlex last updated on 31/Jul/20

$${cooll}\:\&\:{jooss} \\$$