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Question Number 41707 by Necxx last updated on 11/Aug/18
Here′s a question that has troubled me for months now.Please help 5^(√x) − 5^(x−7) =100 find the possible value(s) of x.
$${Here}'{s}\:{a}\:{question}\:{that}\:{has} \\ $$$${troubled}\:{me}\:{for}\:{months}\:{now}.{Please} \\ $$$${help} \\ $$$$ \\ $$$$\mathrm{5}^{\sqrt{{x}}} \:−\:\mathrm{5}^{{x}−\mathrm{7}} \:=\mathrm{100} \\ $$$$ \\ $$$${find}\:{the}\:{possible}\:{value}\left({s}\right)\:{of}\:{x}. \\ $$
Answered by alex041103 last updated on 11/Aug/18
let f(x)=5^(√x) −5^(x−7) We wantf(x)>0 ⇒5^(√x) >5^(x−7) ⇒(√x)>x−7 let u=(√x), u>0 ⇒u^2 −u−7<0 for u∈(((1−(√(29)))/2), ((1+(√(29)))/2))∪(0,∞) ⇒(√x)∈(0, ((1+(√(29)))/2)) ⇒x∈(0,((15+(√(29)))/2))≈(0, 10.19) ⇒solutions will be between 0 and 10.19 We can immediatly see that x=9 is a solution. We see that f ′(x)=ln(5)((5^(√x) /(2(√x)))−5^(x−7) ). We see that f ′(9)<0⇒the function has values biger than 100. By initializing numerical methods we see that x≈8.73 is a solution. By initializing the graph of f we see that there are 2 solutions. Ans. x=9 and x≈8.73
$${let}\:{f}\left({x}\right)=\mathrm{5}^{\sqrt{{x}}} −\mathrm{5}^{{x}−\mathrm{7}} \\ $$$${We}\:{wantf}\left({x}\right)>\mathrm{0} \\ $$$$\Rightarrow\mathrm{5}^{\sqrt{{x}}} >\mathrm{5}^{{x}−\mathrm{7}} \\ $$$$\Rightarrow\sqrt{{x}}>{x}−\mathrm{7} \\ $$$${let}\:{u}=\sqrt{{x}},\:{u}>\mathrm{0} \\ $$$$\Rightarrow{u}^{\mathrm{2}} −{u}−\mathrm{7}<\mathrm{0} \\ $$$$\:{for}\:{u}\in\left(\frac{\mathrm{1}−\sqrt{\mathrm{29}}}{\mathrm{2}},\:\frac{\mathrm{1}+\sqrt{\mathrm{29}}}{\mathrm{2}}\right)\cup\left(\mathrm{0},\infty\right) \\ $$$$\Rightarrow\sqrt{{x}}\in\left(\mathrm{0},\:\frac{\mathrm{1}+\sqrt{\mathrm{29}}}{\mathrm{2}}\right) \\ $$$$\Rightarrow{x}\in\left(\mathrm{0},\frac{\mathrm{15}+\sqrt{\mathrm{29}}}{\mathrm{2}}\right)\approx\left(\mathrm{0},\:\mathrm{10}.\mathrm{19}\right) \\ $$$$\Rightarrow{solutions}\:{will}\:{be}\:{between}\:\mathrm{0}\:{and}\:\mathrm{10}.\mathrm{19} \\ $$$${We}\:{can}\:{immediatly}\:{see}\:{that} \\ $$$${x}=\mathrm{9}\:{is}\:{a}\:{solution}. \\ $$$${We}\:{see}\:{that}\:{f}\:'\left({x}\right)={ln}\left(\mathrm{5}\right)\left(\frac{\mathrm{5}^{\sqrt{{x}}} }{\mathrm{2}\sqrt{{x}}}−\mathrm{5}^{{x}−\mathrm{7}} \right). \\ $$$${We}\:{see}\:{that}\:{f}\:'\left(\mathrm{9}\right)<\mathrm{0}\Rightarrow{the}\:{function} \\ $$$${has}\:{values}\:{biger}\:{than}\:\mathrm{100}. \\ $$$${By}\:{initializing}\:{numerical}\:{methods} \\ $$$${we}\:{see}\:{that}\:{x}\approx\mathrm{8}.\mathrm{73}\:{is}\:{a}\:{solution}. \\ $$$${By}\:{initializing}\:{the}\:{graph}\:{of}\:{f} \\ $$$${we}\:{see}\:{that}\:{there}\:{are}\:\mathrm{2}\:{solutions}. \\ $$$${Ans}.\:{x}=\mathrm{9}\:{and}\:{x}\approx\mathrm{8}.\mathrm{73} \\ $$
Commented by alex041103 last updated on 11/Aug/18
Commented by alex041103 last updated on 11/Aug/18
for example− Newton′s method
$${for}\:{example}−\:{Newton}'{s}\:{method} \\ $$
Commented by Necxx last updated on 11/Aug/18
which numerical method did you apply?
$${which}\:{numerical}\:{method}\:{did}\:{you} \\ $$$${apply}? \\ $$
Commented by Necxx last updated on 11/Aug/18
thanks boss
$${thanks}\:{boss} \\ $$
Answered by peter frank last updated on 04/Oct/18
5^(√x) −5^(x−7) =125−25=5^3 −5^2 5^(√x) =5^3 .⇒x=9 5^(x−7) =5^2 ⇒x=9
$$ \\ $$$$\mathrm{5}^{\sqrt{\mathrm{x}}} −\mathrm{5}^{\mathrm{x}−\mathrm{7}} =\mathrm{125}−\mathrm{25}=\mathrm{5}^{\mathrm{3}} −\mathrm{5}^{\mathrm{2}} \\ $$$$\mathrm{5}^{\sqrt{\mathrm{x}}} =\mathrm{5}^{\mathrm{3}} .\Rightarrow\mathrm{x}=\mathrm{9} \\ $$$$\mathrm{5}^{\mathrm{x}−\mathrm{7}} =\mathrm{5}^{\mathrm{2}} \Rightarrow\mathrm{x}=\mathrm{9} \\ $$$$ \\ $$

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