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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].)

$$\mathrm{Suppose}\:{x}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{number} \\ $$ $$\mathrm{such}\:\mathrm{that}\:\left\{{x}\right\},\:\left[{x}\right]\:\mathrm{and}\:{x}\:\mathrm{are}\:\mathrm{in}\:\mathrm{a} \\ $$ $$\mathrm{geometric}\:\mathrm{progression}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{least} \\ $$ $$\mathrm{positive}\:\mathrm{integer}\:{n}\:\mathrm{such}\:\mathrm{that}\:{x}^{{n}} \:>\:\mathrm{100}. \\ $$ $$\left(\mathrm{Here}\:\left[{x}\right]\:\mathrm{denotes}\:\mathrm{the}\:\mathrm{integer}\:\mathrm{part}\:\mathrm{of}\:{x}\right. \\ $$ $$\left.\mathrm{and}\:\left\{{x}\right\}\:=\:{x}\:−\:\left[{x}\right].\right) \\ $$

Commented byajfour last updated on 30/Aug/17

Is it n>(2/(log _(10) ((((√5)+1)/2)))) ⇒ (n)_(min) =10 .

$${Is}\:{it}\:{n}>\frac{\mathrm{2}}{\mathrm{log}\:_{\mathrm{10}} \left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\right)} \\ $$ $$\Rightarrow\:\:\:\left({n}\right)_{{min}} =\mathrm{10}\:. \\ $$

Commented byTinkutara last updated on 30/Aug/17

Yes. But can it be solved in an exam where calculators and log tables are not allowed? I know the solution so no need to post it.

$$\mathrm{Yes}.\:\mathrm{But}\:\mathrm{can}\:\mathrm{it}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{in}\:\mathrm{an}\:\mathrm{exam} \\ $$ $$\mathrm{where}\:\mathrm{calculators}\:\mathrm{and}\:\mathrm{log}\:\mathrm{tables}\:\mathrm{are} \\ $$ $$\mathrm{not}\:\mathrm{allowed}?\:\mathrm{I}\:\mathrm{know}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{so}\:\mathrm{no} \\ $$ $$\mathrm{need}\:\mathrm{to}\:\mathrm{post}\:\mathrm{it}. \\ $$

Commented byajfour last updated on 30/Aug/17

log ((((√5)+1)/2))=log (((2.236+1)/2)) =log (1.618)=log (16.18)−1 ≈4(0.3010)−1 ≈ 0.2 (2/(log ((((√5)+1)/2)))) ≈ 10 .

$$\mathrm{log}\:\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\right)=\mathrm{log}\:\left(\frac{\mathrm{2}.\mathrm{236}+\mathrm{1}}{\mathrm{2}}\right) \\ $$ $$=\mathrm{log}\:\left(\mathrm{1}.\mathrm{618}\right)=\mathrm{log}\:\left(\mathrm{16}.\mathrm{18}\right)−\mathrm{1} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\approx\mathrm{4}\left(\mathrm{0}.\mathrm{3010}\right)−\mathrm{1}\:\approx\:\mathrm{0}.\mathrm{2} \\ $$ $$\frac{\mathrm{2}}{\mathrm{log}\:\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\right)}\:\approx\:\mathrm{10}\:. \\ $$

Commented byTinkutara last updated on 31/Aug/17

Thanks!

$$\mathrm{Thanks}! \\ $$

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