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Category: Algebra

If-log-logx-3log-logx-log-log-logx-27-Find-x-

Question Number 151013 by mathdanisur last updated on 17/Aug/21 $$\mathrm{If}\:\:\mathrm{log}\left(\mathrm{log}\boldsymbol{\mathrm{x}}\right)^{\frac{\mathrm{3}\boldsymbol{\mathrm{log}}\left(\boldsymbol{\mathrm{logx}}\right)}{\boldsymbol{\mathrm{log}}\left(\boldsymbol{\mathrm{log}}\left(\boldsymbol{\mathrm{logx}}\right)\right)}\:} \:=\:\mathrm{27} \\ $$$$\mathrm{Find}\:\:\boldsymbol{\mathrm{x}}=? \\ $$ Answered by Olaf_Thorendsen last updated on 17/Aug/21 $$\mathrm{log}\left(\mathrm{log}{x}\right)^{\frac{\mathrm{3log}\left(\mathrm{log}{x}\right)}{\mathrm{log}\left(\mathrm{log}\left(\mathrm{log}{x}\right)\right)}} \:=\:\mathrm{27}\:=\:\mathrm{3}^{\mathrm{3}} \\…

If-and-are-the-roots-of-equation-x-2-px-q-0-and-2-2-are-roots-of-the-equation-x-2-rx-s-0-show-that-the-equation-x-2-4qx-2q-2-r-0-has-real-roots-

Question Number 19945 by Tinkutara last updated on 18/Aug/17 $$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{equation} \\ $$$${x}^{\mathrm{2}} \:+\:{px}\:+\:{q}\:=\:\mathrm{0}\:\mathrm{and}\:\alpha^{\mathrm{2}} ,\:\beta^{\mathrm{2}} \:\mathrm{are}\:\mathrm{roots}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{2}} \:−\:{rx}\:+\:{s}\:=\:\mathrm{0},\:\mathrm{show} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{2}} \:−\:\mathrm{4}{qx}\:+\:\mathrm{2}{q}^{\mathrm{2}} \:−\:{r}\:=\:\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{real}\:\mathrm{roots}. \\…

if-x-4-1-3-2-1-3-1-find-3-x-3-x-2-1-x-3-

Question Number 150979 by mathdanisur last updated on 17/Aug/21 $$\mathrm{if}\:\:\mathrm{x}=\sqrt[{\mathrm{3}}]{\mathrm{4}}+\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\:\:\mathrm{find}\:\:\frac{\mathrm{3}}{\mathrm{x}}+\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }=? \\ $$$$ \\ $$ Answered by john_santu last updated on 18/Aug/21 $$\:\mathrm{If}\:\mathrm{x}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{4}}+\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\:\mathrm{then}\:\frac{\mathrm{3}}{\mathrm{x}}+\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}}…

Question-150968

Question Number 150968 by mathdanisur last updated on 17/Aug/21 Answered by dumitrel last updated on 17/Aug/21 $$\Sigma\frac{\left(\mathrm{2}{x}+{y}\right)\left(\mathrm{3}{x}^{\mathrm{2}} +{xy}+\mathrm{2}{y}^{\mathrm{2}} \right)}{\mathrm{2}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\leqslant\Sigma\mathrm{3}\left({x}+{y}\right)=\mathrm{6}\left({x}+{y}+{z}\right) \\ $$ Commented by…

Prove-that-this-is-an-identity-in-x-x-a-x-b-c-a-c-b-x-b-x-c-a-b-a-c-x-c-x-a-b-c-b-a-1-

Question Number 19900 by Tinkutara last updated on 17/Aug/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{this}\:\mathrm{is}\:\mathrm{an}\:\mathrm{identity}\:\mathrm{in}\:{x}: \\ $$$$\frac{\left({x}−{a}\right)\left({x}−{b}\right)}{\left({c}−{a}\right)\left({c}−{b}\right)}+\frac{\left({x}−{b}\right)\left({x}−{c}\right)}{\left({a}−{b}\right)\left({a}−{c}\right)}+\frac{\left({x}−{c}\right)\left({x}−{a}\right)}{\left({b}−{c}\right)\left({b}−{a}\right)}=\mathrm{1} \\ $$ Answered by Rasheed.Sindhi last updated on 17/Aug/17 $$−\frac{\left({x}−{a}\right)\left({x}−{b}\right)}{\left({c}−{a}\right)\left(\mathrm{b}−\mathrm{c}\right)}−\frac{\left({x}−{b}\right)\left({x}−{c}\right)}{\left({a}−{b}\right)\left(\mathrm{c}−\mathrm{a}\right)}−\frac{\left({x}−{c}\right)\left({x}−{a}\right)}{\left({b}−{c}\right)\left(\mathrm{a}−\mathrm{b}\right)}=\mathrm{1} \\ $$$$ \\…

Simplify-i-log-x-i-x-i-

Question Number 19898 by Tinkutara last updated on 17/Aug/17 $$\mathrm{Simplify}\::\:{i}\:\mathrm{log}\:\left(\frac{{x}\:−\:{i}}{{x}\:+\:{i}}\right). \\ $$ Answered by ajfour last updated on 22/Aug/17 $${let}\:{t}={i}\mathrm{ln}\:\left(\frac{{x}−{i}}{{x}+{i}}\right) \\ $$$${and}\:{let}\:{z}=\frac{{x}−{i}}{{x}+{i}}\: \\ $$$$\Rightarrow\:\mid{z}\mid=\mathrm{1} \\…

if-a-b-N-then-determine-all-the-prime-numbers-p-which-satisfy-p-2-a-p-2-b-

Question Number 150966 by mathdanisur last updated on 17/Aug/21 $$\mathrm{if}\:\:\mathrm{a};\mathrm{b}\in\mathbb{N}^{+} \:\:\mathrm{then}\:\mathrm{determine}\:\mathrm{all}\:\mathrm{the} \\ $$$$\mathrm{prime}\:\mathrm{numbers}\:\boldsymbol{\mathrm{p}}\:\mathrm{which}\:\mathrm{satisfy} \\ $$$$\left(\mathrm{p}\:+\:\mathrm{2}\right)^{\boldsymbol{\mathrm{a}}} \:=\:\left(\mathrm{p}\:-\:\mathrm{2}\right)^{\boldsymbol{\mathrm{b}}} \\ $$ Commented by Rasheed.Sindhi last updated on 17/Aug/21…