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log-2-x-log-3-x-1-5-x-

Question Number 215089 by golsendro last updated on 28/Dec/24 $$\:\:\mathrm{log}\:_{\mathrm{2}} \:\mathrm{x}\:+\:\mathrm{log}\:_{\mathrm{3}} \:\left(\mathrm{x}+\mathrm{1}\right)\:=\:\mathrm{5}\: \\ $$$$\:\mathrm{x}\:=\:? \\ $$ Answered by efronzo1 last updated on 28/Dec/24 $$\:\:\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}+\mathrm{1}\right)−\mathrm{2}\:=\:\mathrm{3}−\mathrm{log}\:_{\mathrm{2}}…

n-0-1-n-2-n-1-2n-1-p-q-p-q-p-q-

Question Number 215091 by mnjuly1970 last updated on 28/Dec/24 $$ \\ $$$$\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \:\frac{\:\Gamma^{\:\mathrm{2}} \left({n}+\mathrm{1}\right)}{\Gamma\:\left(\mathrm{2}{n}+\mathrm{1}\right)}\:=\:? \\ $$$$\:\:\:\:\:\:−−− \\ $$$$\:\:\:\:\beta\:\left({p}\:,{q}\:\right)\:=\:\frac{\:\Gamma\:\left({p}\right)\Gamma\left({q}\right)}{\Gamma\left({p}+{q}\:\right)} \\ $$ Answered by MrGaster…

Question-215052

Question Number 215052 by Abdullahrussell last updated on 27/Dec/24 Answered by mr W last updated on 27/Dec/24 $${f}\left(\mathrm{1}\right)=\mathrm{1}\:\Rightarrow{f}\left({x}\right)=\left({x}−\mathrm{1}\right){g}\left({x}\right)+\mathrm{1} \\ $$$${f}\left(\mathrm{2}\right)=\left(\mathrm{2}−\mathrm{1}\right){g}\left(\mathrm{2}\right)+\mathrm{1}=\mathrm{4}\:\Rightarrow{g}\left(\mathrm{2}\right)=\mathrm{3}\:\Rightarrow{g}\left({x}\right)=\left({x}−\mathrm{2}\right){h}\left({x}\right)+\mathrm{3} \\ $$$$\Rightarrow{f}\left({x}\right)=\left({x}−\mathrm{1}\right)\left[\left({x}−\mathrm{2}\right){h}\left({x}\right)+\mathrm{3}\right]+\mathrm{1} \\ $$$${f}\left(\mathrm{3}\right)=\left(\mathrm{3}−\mathrm{1}\right)\left[\left(\mathrm{3}−\mathrm{2}\right){h}\left(\mathrm{3}\right)+\mathrm{3}\right]+\mathrm{1}=\mathrm{3}\:\Rightarrow{h}\left(\mathrm{3}\right)=−\mathrm{2}\:\Rightarrow{h}\left({x}\right)=\left({x}−\mathrm{3}\right){k}\left({x}\right)−\mathrm{2} \\…

Solve-for-a-b-c-R-1-a-1-b-c-1-2-1-b-1-c-a-1-3-1-c-1-a-b-1-4-

Question Number 215048 by Hanuda354 last updated on 27/Dec/24 $$\mathrm{Solve}\:\:\mathrm{for}\:\:{a},\:{b},\:{c}\:\in\:\mathbb{R} \\ $$$$\:\:\:\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}+{c}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}+{a}}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\frac{\mathrm{1}}{{c}}\:+\:\frac{\mathrm{1}}{{a}+{b}}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$ Commented by Abdullahrussell last updated on 27/Dec/24…

lim-n-k-1-n-k-2-1-k-2-4-

Question Number 215051 by MrGaster last updated on 27/Dec/24 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{{k}^{\mathrm{2}} +\mathrm{1}}{\:\sqrt{{k}^{\mathrm{2}} +\mathrm{4}}}=? \\ $$$$ \\ $$ Answered by MathematicalUser2357 last…

x-2-2000x-1-0-Roots-a-and-b-x-2-2008x-1-0-Roots-c-and-d-Find-a-c-b-d-a-d-b-c-

Question Number 215072 by hardmath last updated on 27/Dec/24 $$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{2000x}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{Roots}:\:\:\boldsymbol{\mathrm{a}}\:\:\mathrm{and}\:\:\boldsymbol{\mathrm{b}} \\ $$$$\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{2008x}\:−\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{Roots}:\:\:\boldsymbol{\mathrm{c}}\:\:\mathrm{and}\:\:\boldsymbol{\mathrm{d}} \\ $$$$\mathrm{Find}:\:\:\left(\mathrm{a}+\mathrm{c}\right)\left(\mathrm{b}+\mathrm{d}\right)\left(\mathrm{a}−\mathrm{d}\right)\left(\mathrm{b}−\mathrm{c}\right)\:=\:? \\ $$ Commented by TonyCWX08…