Question Number 172028 by Mikenice last updated on 23/Jun/22 $${solve}: \\ $$$$\frac{{log}_{\mathrm{2}} \left(\mathrm{9}−\mathrm{2}^{\left.{x}\right)} \right.}{{log}_{\mathrm{2}} \mathrm{2}^{\left(\mathrm{3}−{x}\right)} }={log}_{\mathrm{2}} \mathrm{2} \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 171990 by Mikenice last updated on 22/Jun/22 $${solve}: \\ $$$${log}_{\mathrm{7}} \mathrm{2}\:+\:{log}_{\mathrm{49}} {x}\:={log}_{\mathrm{7}} \sqrt{\mathrm{3}} \\ $$ Answered by nurtani last updated on 23/Jun/22 $$\Leftrightarrow\:{log}_{\mathrm{7}}…
Question Number 40873 by scientist last updated on 28/Jul/18 $${If}\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\mathrm{0},\:\:{prove}\:{that}\:\mathrm{log}\:\left({a}+{b}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{log}\:{a}\:+\mathrm{log}\:{b}\:+\mathrm{log}\:\mathrm{3}\right) \\ $$$$\left[{given}\:{a}+{b}\neq\mathrm{0}\right] \\ $$ Answered by math1967 last updated on 29/Jul/18 $${a}^{\mathrm{3}} +{b}^{\mathrm{3}}…
Question Number 40872 by scientist last updated on 28/Jul/18 $${If}\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\mathrm{0},\:\:{prove}\:{that}\:\mathrm{log}\:\left({a}+{b}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{log}\:{a}\:+\mathrm{log}\:{b}\:+\mathrm{log}\:\mathrm{3}\right) \\ $$$$\left[{given}\:{a}+{b}\neq\mathrm{0}\right] \\ $$ Commented by MrW3 last updated on 29/Jul/18 $${I}\:{don}'{t}\:{think}\:{there}\:{are}\:{such}\:{real}\:{values} \\…
Question Number 106024 by john santu last updated on 02/Aug/20 $$\mathrm{log}\:_{\mathrm{4}} \left(\mathrm{5x}−\mathrm{6}\right).\mathrm{log}\:_{\mathrm{x}} \left(\mathrm{256}\right)=\mathrm{8} \\ $$ Commented by Dwaipayan Shikari last updated on 02/Aug/20 $$\left\{\mathrm{2},\mathrm{3}\right\} \\…
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Question Number 171284 by Kodjo last updated on 11/Jun/22 $${g}\left({x}\right)=−{x}^{\mathrm{2}} +\mathrm{1}−{ln}\mid{x}\mid \\ $$Study the variations of the function g and draw up its table of…
Question Number 171031 by cortano1 last updated on 06/Jun/22 Answered by a.lgnaoui last updated on 07/Jun/22 $$={Log}\left(\frac{{a}}{\mathrm{1}−{b}}\right)+{Log}\left(\frac{\mathrm{1}−{b}}{\mathrm{1}+{b}}\right)={Log}\left(\frac{{a}}{\mathrm{1}−{b}}×\frac{\mathrm{1}−{b}}{\mathrm{1}+{b}}\right)={Log}\left(\frac{{a}}{\mathrm{1}+{b}}\right) \\ $$$$={Log}\left(\frac{{a}−{b}+\mathrm{1}}{{a}+{b}+\mathrm{1}}\right)^{\mathrm{3}} =\mathrm{3}{Log}\left(\frac{{a}+\mathrm{1}−{b}}{{a}+\mathrm{1}+{b}}\right)=\mathrm{3}{Log}\left(\frac{\left({a}+\mathrm{1}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} }{\left({a}+\mathrm{1}+{b}\right)^{\mathrm{2}} }\right) \\ $$$$=\mathrm{3}{Log}\left[\frac{\left({a}+\mathrm{1}\right)^{\mathrm{2}}…
Question Number 170958 by cortano1 last updated on 04/Jun/22 $$\:\:\:\:\:\:\mathrm{log}\:_{\mathrm{5}} \left({x}+\mathrm{3}\right)=\mathrm{log}\:_{\mathrm{6}} \left({x}+\mathrm{14}\right) \\ $$ Answered by floor(10²Eta[1]) last updated on 05/Jun/22 $$\mathrm{log}_{\mathrm{5}} \left(\mathrm{x}+\mathrm{3}\right)=\mathrm{y}=\mathrm{log}_{\mathrm{6}} \left(\mathrm{x}+\mathrm{14}\right) \\…