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

solve 7^x  + 24^x  = 25^x

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

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

((7/(25)))^x +(((24)/(25)))^x =1  Let (7/(25))=sin θ,     cos θ=(√(1−sin^2 θ))=(√(1−((7/(25)))^2 ))         =(√(1−((49)/(625)))) =(√((625−49)/(625)))=((24)/(25))  ∴ ((7/(25)))^x +(((24)/(25)))^x =1⇒                        (sin θ)^x +(cos θ)^x =1  Comparing with the identity:                       (sin θ)^2 +(cos θ)^2 =1                        x=2

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

Commented by bramlex last updated on 31/Jul/20

cooll & jooss

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

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