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Category: Logarithms

Question-99681

Question Number 99681 by Algoritm last updated on 22/Jun/20 Answered by floor(10²Eta[1]) last updated on 22/Jun/20 $$\mathrm{1}−\mathrm{2}{x}>\mathrm{0}\Rightarrow{x}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{1}−\mathrm{2}{x}\neq\mathrm{1}\Rightarrow{x}\neq\mathrm{0} \\ $$$${but}\:{we}\:{know}\:{that}\:{for}\:{all}\:{x}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}−\mathrm{2018}\:{will}\:{be}\:{negative},\:{so}\:{for}\:{that} \\ $$$${expression}\:{be}\:{positive}…

x-R-log-2-x-2-3-log-2-2x-3-28-

Question Number 165178 by cortano1 last updated on 27/Jan/22 $$\:\:{x}\in{R}\:\Rightarrow\:\mid\:\mathrm{log}\:_{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\mid^{\mathrm{3}} +\mid\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{2}{x}\right)\mid^{\mathrm{3}} =\mathrm{28} \\ $$ Answered by aleks041103 last updated on 27/Jan/22 $${log}_{\mathrm{2}} \left({x}/\mathrm{2}\right)={log}_{\mathrm{2}}…

Question-165068

Question Number 165068 by cortano1 last updated on 25/Jan/22 Commented by bobhans last updated on 25/Jan/22 $$\:\Rightarrow\mathrm{x}^{\mathrm{log}\:_{\mathrm{81}} \left(\mathrm{243}\right)} −\mathrm{2x}\:=\:\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{log}\:_{\mathrm{16}} \left(\mathrm{2}\right)} −\mathrm{8} \\ $$$$\Rightarrow\mathrm{x}^{\frac{\mathrm{5}}{\mathrm{4}}} −\mathrm{2x}\:=\:\left(\mathrm{x}+\mathrm{2}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} −\mathrm{8}…

Question-164985

Question Number 164985 by cortano1 last updated on 24/Jan/22 Answered by Eulerian last updated on 24/Jan/22 $$\: \\ $$$$\:\boldsymbol{\mathrm{Solution}}: \\ $$$$\:\mathrm{3}\centerdot\frac{\mathrm{log}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{log}^{\mathrm{2}} \left(\mathrm{2}\right)}\:−\:\mathrm{1}\:−\:\mathrm{9}\centerdot\frac{\mathrm{log}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{log}^{\mathrm{2}} \left(\mathrm{x}\right)}\:=\:\mathrm{25}…

Question-164806

Question Number 164806 by saboorhalimi last updated on 22/Jan/22 Commented by cortano1 last updated on 22/Jan/22 $$\:\mathrm{log}\:_{\mathrm{12}} \left(\mathrm{3}\right)={a}\:\Rightarrow\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{12}\right)=\frac{\mathrm{1}}{{a}} \\ $$$$\Rightarrow\mathrm{1}+\mathrm{2}\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)=\frac{\mathrm{1}}{{a}}\: \\ $$$$\Rightarrow\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)=\:\frac{\mathrm{1}−{a}}{\mathrm{2}{a}}\:…

Question-32885

Question Number 32885 by scientist last updated on 05/Apr/18 Answered by MJS last updated on 05/Apr/18 $$\mathrm{log}\:{u}={a} \\ $$$$\mathrm{log}\:{v}={b} \\ $$$$\mathrm{log}\:{s}={c} \\ $$$$\mathrm{log}\:{t}={d} \\ $$$$\mathrm{all}\:\mathrm{we}\:\mathrm{need}\:\mathrm{is}\:\mathrm{log}\:\frac{{p}}{{q}}=\mathrm{log}\:{p}−\mathrm{log}\:{q}…