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Question Number 6483 by sanusihammed last updated on 28/Jun/16

5^(√(x )) − 5^((x − 7)) = 100 Find the value of x

$$\mathrm{5}^{\sqrt{{x}\:}} \:\:−\:\:\:\mathrm{5}^{\left({x}\:−\:\mathrm{7}\right)} \:\:=\:\:\mathrm{100} \\ $$$$ \\ $$$${Find}\:{the}\:{value}\:{of}\:{x} \\ $$

Answered by Yozzii last updated on 29/Jun/16

5^(√x) −5^(−7) 5^x =100 Let u=(√x)⇒u^2 =x ∴ 5^u −5^(−7) 5^u^2 =100 5^u −5^(u^2 −7) =100=2^2 5^2 For both l.h.s and r.h.s to be integers (100) then u^2 −7≥0⇒u^2 ≥7. Try u=3 if u∈Z. ⇒u^2 ≥7 and 5^3 −5^(9−7) =125−25=100. So, x=u^2 =9 satisfies the given equation. There possibly exists other values of x since this does not provide proof of x=9 as the only answer. If l.h.s>0, as 100>0, we require that u>u^2 −7 u^2 −u−7<0 (u−((1+(√(29)))/2))(u−((1−(√(29)))/2))<0 ⇒((1−(√(29)))/2)<u<((1+(√(29)))/2) ((1−(√(29)))/2)<(√x)<((1+(√(29)))/2) (((1−(√(29)))/2))^2 <x<(((1+(√(29)))/2))^2 4.807...<x<10.19... 5^(√x) −5^(x−7) =100. 5^((√x)+7−x) −1=2^2 5^(9−x)

$$\mathrm{5}^{\sqrt{{x}}} −\mathrm{5}^{−\mathrm{7}} \mathrm{5}^{{x}} =\mathrm{100} \\ $$$${Let}\:{u}=\sqrt{{x}}\Rightarrow{u}^{\mathrm{2}} ={x} \\ $$$$\therefore\:\mathrm{5}^{{u}} −\mathrm{5}^{−\mathrm{7}} \mathrm{5}^{{u}^{\mathrm{2}} } =\mathrm{100} \\ $$$$\mathrm{5}^{{u}} −\mathrm{5}^{{u}^{\mathrm{2}} −\mathrm{7}} =\mathrm{100}=\mathrm{2}^{\mathrm{2}} \mathrm{5}^{\mathrm{2}} \\ $$$${For}\:{both}\:{l}.{h}.{s}\:{and}\:{r}.{h}.{s}\:{to}\:{be}\:{integers}\:\left(\mathrm{100}\right)\: \\ $$$${then}\:{u}^{\mathrm{2}} −\mathrm{7}\geqslant\mathrm{0}\Rightarrow{u}^{\mathrm{2}} \geqslant\mathrm{7}.\:{Try}\:{u}=\mathrm{3}\:{if}\:{u}\in\mathbb{Z}. \\ $$$$\Rightarrow{u}^{\mathrm{2}} \geqslant\mathrm{7}\:{and}\:\mathrm{5}^{\mathrm{3}} −\mathrm{5}^{\mathrm{9}−\mathrm{7}} =\mathrm{125}−\mathrm{25}=\mathrm{100}. \\ $$$${So},\:{x}={u}^{\mathrm{2}} =\mathrm{9}\:{satisfies}\:{the}\:{given}\:{equation}. \\ $$$${There}\:{possibly}\:{exists}\:{other}\:{values}\:{of} \\ $$$${x}\:{since}\:{this}\:{does}\:{not}\:{provide}\:{proof}\:{of} \\ $$$${x}=\mathrm{9}\:{as}\:{the}\:{only}\:{answer}. \\ $$$${If}\:{l}.{h}.{s}>\mathrm{0},\:{as}\:\mathrm{100}>\mathrm{0},\:{we}\:{require}\:{that} \\ $$$${u}>{u}^{\mathrm{2}} −\mathrm{7} \\ $$$${u}^{\mathrm{2}} −{u}−\mathrm{7}<\mathrm{0} \\ $$$$\left({u}−\frac{\mathrm{1}+\sqrt{\mathrm{29}}}{\mathrm{2}}\right)\left({u}−\frac{\mathrm{1}−\sqrt{\mathrm{29}}}{\mathrm{2}}\right)<\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}−\sqrt{\mathrm{29}}}{\mathrm{2}}<{u}<\frac{\mathrm{1}+\sqrt{\mathrm{29}}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}−\sqrt{\mathrm{29}}}{\mathrm{2}}<\sqrt{{x}}<\frac{\mathrm{1}+\sqrt{\mathrm{29}}}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{1}−\sqrt{\mathrm{29}}}{\mathrm{2}}\right)^{\mathrm{2}} <{x}<\left(\frac{\mathrm{1}+\sqrt{\mathrm{29}}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}.\mathrm{807}...<{x}<\mathrm{10}.\mathrm{19}... \\ $$$$\mathrm{5}^{\sqrt{{x}}} −\mathrm{5}^{{x}−\mathrm{7}} =\mathrm{100}. \\ $$$$\mathrm{5}^{\sqrt{{x}}+\mathrm{7}−{x}} −\mathrm{1}=\mathrm{2}^{\mathrm{2}} \mathrm{5}^{\mathrm{9}−{x}} \\ $$$$ \\ $$

Commented by sanusihammed last updated on 29/Jun/16

Thanks so much

$${Thanks}\:{so}\:{much} \\ $$

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