Question Number 18034 by chux last updated on 14/Jul/17 $$\mathrm{solve}\:\mathrm{for}\:\mathrm{a} \\ $$$$ \\ $$$$\mathrm{5log}\:_{\mathrm{4}} \mathrm{a}+\mathrm{48log}\:_{\mathrm{a}} \mathrm{4}=\frac{\mathrm{a}}{\mathrm{8}} \\ $$ Answered by 433 last updated on 14/Jul/17…
Question Number 18022 by ibraheem160 last updated on 13/Jul/17 $${The}\:{first}\:{term}\:{of}\:{an}\:{A}.{P}\:\:{is}\:{log}^{{a}} \:{and}\:{second}\:{term}\:{is}\: \\ $$$${log}^{{b}} .{show}\:{that}\:{the}\:{sum}\:{of}\:{first}\:{n}\:{terms}\:{in}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{log}\left[\frac{{b}^{{n}\left({n}−\mathrm{1}\right)} }{{a}^{{n}\left(−\mathrm{3}\right)} }\right] \\ $$ Answered by prakash jain last…
Question Number 17748 by ibraheem160 last updated on 10/Jul/17 $${f}\:{p}^{\mathrm{2}} ={qr},\:\:{prove}\:{thathat}\:{log}_{{r}} ^{{p}} +{log}_{{q}=} ^{{p}} \\ $$$$\mathrm{2}{log}_{{q}} ^{{p}} {log}_{{r}} ^{{p}} \\ $$ Commented by tawa tawa…
Question Number 148494 by liberty last updated on 28/Jul/21 $$\:\:\:\frac{\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2008}} }{\left(\mathrm{7}+\mathrm{5}\sqrt{\mathrm{2}}\right)^{\mathrm{1338}} }\:+\:\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)\:=\:\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{x}\right) \\ $$$$\:\mathrm{x}=?\: \\ $$ Answered by EDWIN88 last updated on 28/Jul/21 $$\:\mathrm{log}\:_{\mathrm{2}}…
Question Number 17386 by ajfour last updated on 05/Jul/17 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mathrm{log}\:_{\mathrm{2}} \mathrm{x}\:\mathrm{log}\:_{\mathrm{3}} \mathrm{x}\:\mathrm{log}\:_{\mathrm{5}} \mathrm{x}=\mathrm{log}\:_{\mathrm{2}} \mathrm{x}\:\mathrm{log}\:_{\mathrm{3}} \mathrm{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{log}\:_{\mathrm{3}} \mathrm{x}\:\mathrm{log}\:_{\mathrm{5}} \mathrm{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{log}\:_{\mathrm{5}} \mathrm{x}\:\mathrm{log}\:_{\mathrm{2}} \mathrm{x}\:.…
Question Number 148312 by puissant last updated on 27/Jul/21 $${calculer}\:{la}\:{differentielle}\:{de}\: \\ $$$${y}={log}\left({x}\right) \\ $$$${teste}:\:{sachant}\:{que}\:{log}\left(\mathrm{35}\right)=\mathrm{1},\mathrm{54407}, \\ $$$${calculer}\:{log}\left(\mathrm{3501}\right) \\ $$$${NB}:\:{on}\:{rappelle}\:{que}\:\frac{\mathrm{1}}{{log}\left(\mathrm{10}\right)}={log}\left({e}\right)=\mathrm{0},\mathrm{43429}.. \\ $$ Answered by Olaf_Thorendsen last updated…
Question Number 148257 by liberty last updated on 26/Jul/21 Answered by Olaf_Thorendsen last updated on 26/Jul/21 $$\left(\mathrm{2}^{{x}} −\mathrm{4}\right)^{\mathrm{3}} +\left(\mathrm{4}^{{x}} −\mathrm{2}\right)^{\mathrm{3}} \:=\:\left(\mathrm{4}^{{x}} +\mathrm{2}^{{x}} −\mathrm{6}\right)^{\mathrm{3}} \\ $$$$\mathrm{Let}\:{u}\:=\:\mathrm{2}^{{x}}…
Question Number 148242 by puissant last updated on 26/Jul/21 Answered by Jonathanwaweh last updated on 26/Jul/21 $${soit}\:\:{Aappartenant}\:{a}\:\varepsilon\:{on}\:{a}\:{got}\left({A}\right)={g}\left(\overset{\rightarrow} {{A}}+\overset{\rightarrow} {{V}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={g}\left({A}\overset{\rightarrow} {\right)}+{g}\left({V}\overset{\rightarrow} {\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:{de}\:{meme}\:{tog}\left({A}\right)=\overset{\rightarrow}…
Question Number 16973 by tawa tawa last updated on 29/Jun/17 $$\mathrm{5}^{\mathrm{log}\left(\mathrm{x}\right)} \:=\:\mathrm{x}^{\mathrm{log}\left(\mathrm{2}\right)} ,\:\:\:\:\:\mathrm{find}\:\:\mathrm{x}. \\ $$ Answered by mrW1 last updated on 29/Jun/17 $$\mathrm{log}\:\left(\mathrm{x}\right)\:\mathrm{log}\:\left(\mathrm{5}\right)=\mathrm{log}\:\left(\mathrm{2}\right)\:\mathrm{log}\:\left(\mathrm{x}\right) \\ $$$$\mathrm{log}\:\left(\mathrm{x}\right)\left[\:\mathrm{log}\:\left(\mathrm{5}\right)−\mathrm{log}\:\left(\mathrm{2}\right)\right]=\mathrm{0}…
Question Number 82339 by function last updated on 20/Feb/20 Commented by TANMAY PANACEA last updated on 20/Feb/20 $${pls}\:{recheck}\:{the}\:{question} \\ $$ Terms of Service Privacy Policy…