Question Number 221814 by Tawa11 last updated on 10/Jun/25
Answered by wewji12 last updated on 11/Jun/25
$$\mathrm{log}_{{a}} \left({b}\right)=\frac{\mathrm{log}_{{b}} \left({c}\right)}{\mathrm{log}_{{a}} \left({c}\right)} \\ $$$$\frac{\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{ln}\left({x}+\mathrm{1}\right)}+\frac{\mathrm{ln}\left(\mathrm{3}\right)}{\mathrm{ln}\left({x}+\mathrm{2}\right)}+\frac{\mathrm{ln}\left(\mathrm{4}\right)}{\mathrm{ln}\left({x}+\mathrm{3}\right)}=\mathrm{3} \\ $$$$\mathrm{if}\:{x}=\mathrm{1}\: \\ $$$$\frac{\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{ln}\left(\mathrm{2}\right)}+\frac{\mathrm{ln}\left(\mathrm{3}\right)}{\mathrm{ln}\left(\mathrm{3}\right)}+\frac{\mathrm{ln}\left(\mathrm{4}\right)}{\mathrm{ln}\left(\mathrm{4}\right)}=\mathrm{3}. \\ $$$$\therefore\:{x}=\mathrm{1}…. \\ $$
Commented by wewji12 last updated on 11/Jun/25
$$\mathrm{if}\:{x}<\mathrm{0}…?\:\: \\ $$$${x}_{{n}+\mathrm{1}} ={x}_{{n}} −\frac{{f}\left({x}_{{n}} \right)}{{f}\:'\left({x}_{{n}} \right)}\:\:\mathrm{and}\:\mathrm{Let}'\mathrm{s}\:\mathrm{start}\:{x}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{3}}\:\: \\ $$$${f}\left({x}\right)=\mathrm{log}_{{x}+\mathrm{1}} \left(\mathrm{2}\right)+\mathrm{log}_{{x}+\mathrm{2}} \left(\mathrm{3}\right)+\mathrm{log}_{{x}+\mathrm{3}} \left(\mathrm{4}\right)−\mathrm{3} \\ $$$$\mathrm{because}\:\mathrm{we}\:\mathrm{finding}\:\mathrm{value}\:''{x}'' \\ $$$$\:\mathrm{log}_{{x}+\mathrm{1}} \left(\mathrm{2}\right)+\mathrm{log}_{{x}+\mathrm{2}} \left(\mathrm{3}\right)+\mathrm{log}_{{x}+\mathrm{3}} \left(\mathrm{4}\right)=\mathrm{3} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{can}\:\mathrm{set}\:{f}\left({x}\right)\:\mathrm{as} \\ $$$${f}\left({x}\right)=\mathrm{log}_{{x}+\mathrm{1}} \left(\mathrm{2}\right)+\mathrm{log}_{{x}+\mathrm{2}} \left(\mathrm{3}\right)+\mathrm{log}_{{x}+\mathrm{3}} \left(\mathrm{4}\right)−\mathrm{3} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{x}_{{n}+\mathrm{1}} \approx−\mathrm{0}.\mathrm{47190214154884974}….. \\ $$
Commented by MathematicalUser2357 last updated on 11/Jun/25
Did you just forgot the rules of logarithms and derivatives and integrals?
Commented by wewji12 last updated on 11/Jun/25
미분이요? 미분해서 어떻게하실려고요 ㅋㅋㅋㅋㅋ
그리고 각 각 log(x+1,2) log(x+2,3) log(x+3,4)를
곱한다음에 다시 미분을 할려해도 과정이 복잡할뿐이고
왠 미적분....? 알다싶이 a에서의 함수 f(x)의
f(a)+f'(a)(x-a)가 그 함수의 점(a,f(a))에서의
접선의 방정식이고 그걸 이용해서 영점 근사하는것도
뭐 일종의 미적분을 응용한 방법이긴한데 무슨말씀이신
이해가안가네요
Commented by MathematicalUser2357 last updated on 13/Jun/25
then eat this apple and grow your memory
Commented by MathematicalUser2357 last updated on 13/Jun/25
$$\boldsymbol{\mathrm{I}}\:\boldsymbol{\mathrm{was}}\:\boldsymbol{\mathrm{drunk}}\:\boldsymbol{\mathrm{bruh}} \\ $$
Commented by fantastic last updated on 13/Jun/25
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