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Question Number 118203 by aurpeyz last updated on 15/Oct/20

Answered by MJS_new last updated on 15/Oct/20

at first glance x=1

$$\mathrm{at}\:\mathrm{first}\:\mathrm{glance}\:{x}=\mathrm{1} \\ $$

Commented by aurpeyz last updated on 17/Oct/20

workings?

$${workings}? \\ $$

Commented by MJS_new last updated on 17/Oct/20

sorry “at first glance” means, no workings  needed  log_n  n =1  log_(x+1)  2 +log_(x+2)  3 +log_(x+3)  4 =3  x+1=2∧x+2=3∧x+3=4 ⇒ x=1  ⇒  1+1+1=3

$$\mathrm{sorry}\:``\mathrm{at}\:\mathrm{first}\:\mathrm{glance}''\:\mathrm{means},\:\mathrm{no}\:\mathrm{workings} \\ $$$$\mathrm{needed} \\ $$$$\mathrm{log}_{{n}} \:{n}\:=\mathrm{1} \\ $$$$\mathrm{log}_{{x}+\mathrm{1}} \:\mathrm{2}\:+\mathrm{log}_{{x}+\mathrm{2}} \:\mathrm{3}\:+\mathrm{log}_{{x}+\mathrm{3}} \:\mathrm{4}\:=\mathrm{3} \\ $$$${x}+\mathrm{1}=\mathrm{2}\wedge{x}+\mathrm{2}=\mathrm{3}\wedge{x}+\mathrm{3}=\mathrm{4}\:\Rightarrow\:{x}=\mathrm{1} \\ $$$$\Rightarrow \\ $$$$\mathrm{1}+\mathrm{1}+\mathrm{1}=\mathrm{3} \\ $$

Answered by MJS_new last updated on 15/Oct/20

x>−1  ⇒ there could be a 2^(nd)  solution with −1<x<0  approximating I get  x≈−.471902

$${x}>−\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{there}\:\mathrm{could}\:\mathrm{be}\:\mathrm{a}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{solution}\:\mathrm{with}\:−\mathrm{1}<{x}<\mathrm{0} \\ $$$$\mathrm{approximating}\:\mathrm{I}\:\mathrm{get} \\ $$$${x}\approx−.\mathrm{471902} \\ $$

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