Question Number 20652 by Tinkutara last updated on 30/Aug/17
![Suppose x is a positive real number such that {x}, [x] and x are in a geometric progression. Find the least positive integer n such that x^n > 100. (Here [x] denotes the integer part of x and {x} = x − [x].)](https://www.tinkutara.com/question/Q20652.png)
$$\mathrm{Suppose}\:{x}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{number} \\ $$$$\mathrm{such}\:\mathrm{that}\:\left\{{x}\right\},\:\left[{x}\right]\:\mathrm{and}\:{x}\:\mathrm{are}\:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{geometric}\:\mathrm{progression}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{least} \\ $$$$\mathrm{positive}\:\mathrm{integer}\:{n}\:\mathrm{such}\:\mathrm{that}\:{x}^{{n}} \:>\:\mathrm{100}. \\ $$$$\left(\mathrm{Here}\:\left[{x}\right]\:\mathrm{denotes}\:\mathrm{the}\:\mathrm{integer}\:\mathrm{part}\:\mathrm{of}\:{x}\right. \\ $$$$\left.\mathrm{and}\:\left\{{x}\right\}\:=\:{x}\:−\:\left[{x}\right].\right) \\ $$
Commented by ajfour last updated on 30/Aug/17
$${Is}\:{it}\:{n}>\frac{\mathrm{2}}{\mathrm{log}\:_{\mathrm{10}} \left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\Rightarrow\:\:\:\left({n}\right)_{{min}} =\mathrm{10}\:. \\ $$
Commented by Tinkutara last updated on 30/Aug/17
$$\mathrm{Yes}.\:\mathrm{But}\:\mathrm{can}\:\mathrm{it}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{in}\:\mathrm{an}\:\mathrm{exam} \\ $$$$\mathrm{where}\:\mathrm{calculators}\:\mathrm{and}\:\mathrm{log}\:\mathrm{tables}\:\mathrm{are} \\ $$$$\mathrm{not}\:\mathrm{allowed}?\:\mathrm{I}\:\mathrm{know}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{so}\:\mathrm{no} \\ $$$$\mathrm{need}\:\mathrm{to}\:\mathrm{post}\:\mathrm{it}. \\ $$
Commented by ajfour last updated on 30/Aug/17
$$\mathrm{log}\:\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\right)=\mathrm{log}\:\left(\frac{\mathrm{2}.\mathrm{236}+\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\mathrm{log}\:\left(\mathrm{1}.\mathrm{618}\right)=\mathrm{log}\:\left(\mathrm{16}.\mathrm{18}\right)−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\approx\mathrm{4}\left(\mathrm{0}.\mathrm{3010}\right)−\mathrm{1}\:\approx\:\mathrm{0}.\mathrm{2} \\ $$$$\frac{\mathrm{2}}{\mathrm{log}\:\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\right)}\:\approx\:\mathrm{10}\:. \\ $$
Commented by Tinkutara last updated on 31/Aug/17
$$\mathrm{Thanks}! \\ $$