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Category: Logarithms

What-is-the-argument-of-the-complex-numbers-below-i-z-1-e-pi-6-i-ii-z-1-e-pi-6-i-

Question Number 145524 by physicstutes last updated on 05/Jul/21 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{argument}\:\mathrm{of}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{numbers}\:\mathrm{below} \\ $$$$\left(\mathrm{i}\right)\:{z}\:=\:\mathrm{1}+{e}^{\frac{\pi}{\mathrm{6}}{i}} \\ $$$$\left(\mathrm{ii}\right)\:{z}\:=\:\mathrm{1}\:−{e}^{\frac{\pi}{\mathrm{6}}{i}} \\ $$ Answered by Olaf_Thorendsen last updated on 05/Jul/21 $$\left({i}\right)\:{z}\:=\:\mathrm{1}+{e}^{{i}\frac{\pi}{\mathrm{6}}} \\…

original-length-of-the-iron-rod-175-65-increase-6-1-3-175-65-19-3-1-100-175-65-19-175-65-3-100-3337-35-300-11-1245-new-length-original-length-increased-length-175-65-11-1245

Question Number 145437 by Saiki last updated on 04/Jul/21 $${original}\:{length}\:{of}\:{the}\:{iron}\:{rod}=\mathrm{175}.\mathrm{65} \\ $$$$\%\:{increase}=\mathrm{6}\frac{\mathrm{1}}{\mathrm{3}}\%×\mathrm{175}.\mathrm{65} \\ $$$$=\frac{\mathrm{19}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{100}}×\mathrm{175}.\mathrm{65} \\ $$$$=\frac{\mathrm{19}×\mathrm{175}.\mathrm{65}}{\mathrm{3}×\mathrm{100}}=\frac{\mathrm{3337}.\mathrm{35}}{\mathrm{300}}=\mathrm{11}.\mathrm{1245} \\ $$$${new}\:{length}={original}\:{length}+{increased}\:{length} \\ $$$$=\mathrm{175}.\mathrm{65}+\mathrm{11}.\mathrm{1245} \\ $$$$=\mathrm{186}.\mathrm{7745}{cm} \\ $$$${solution}\:{by}\:{CASIO}….. \\…

Question-145108

Question Number 145108 by imjagoll last updated on 02/Jul/21 Answered by liberty last updated on 02/Jul/21 $$\:\sqrt[{\mathrm{1}/\mathrm{4}}]{\:\sqrt[{\mathrm{1}/\mathrm{5}}]{\frac{\mathrm{2}^{{x}+\mathrm{1}} }{\mathrm{32}^{\mathrm{2}{x}−\mathrm{2}} }}}=\:\mathrm{64} \\ $$$$\Rightarrow\:\sqrt[{\mathrm{1}/\mathrm{4}}]{\:\left(\frac{\mathrm{2}^{{x}+\mathrm{1}} }{\mathrm{2}^{\mathrm{10}{x}−\mathrm{10}} }\right)^{\mathrm{5}} }=\:\mathrm{2}^{\mathrm{6}} \\…

log-x-2-x-2-1-log-x-4-x-x-

Question Number 145071 by imjagoll last updated on 02/Jul/21 $$\mathrm{log}\:_{\left(\frac{\mathrm{x}}{\mathrm{2}}\right)} \left(\mathrm{x}+\mathrm{2}\right)\:=\:\mathrm{1}+\:\mathrm{log}\:_{\mathrm{x}} \left(\mathrm{4}−\mathrm{x}\right) \\ $$$$\:\mathrm{x}=? \\ $$ Answered by MJS_new last updated on 02/Jul/21 $$\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\…

log-3-x-1-log-4-x-8-x-

Question Number 145067 by imjagoll last updated on 02/Jul/21 $$\:\mathrm{log}\:_{\mathrm{3}} \left({x}+\mathrm{1}\right)\:=\mathrm{log}\:_{\mathrm{4}} \left({x}+\mathrm{8}\right) \\ $$$$\:{x}=? \\ $$ Answered by bobhans last updated on 02/Jul/21 $$\:\mathrm{x}=\mathrm{8}\:\Rightarrow\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{8}+\mathrm{1}\right)=\mathrm{log}\:_{\mathrm{4}}…

4-2x-1-1-4-2-log-2-2x-gt-2-log-x-2-log-1-x-2-2x-

Question Number 79263 by jagoll last updated on 23/Jan/20 $$\mathrm{4}^{\mathrm{2x}−\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{4}}\:^{\mathrm{2}} \mathrm{log}^{\mathrm{2}} \left(\mathrm{2x}\right)>\:^{\mathrm{2}} \mathrm{log}\left(\mathrm{x}\right) \\ $$$$\left\{^{\mathrm{2}} \mathrm{log}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)−\mathrm{2}^{\mathrm{2x}} \right\} \\ $$ Commented by john santu last…

Question-13151

Question Number 13151 by Tinkutara last updated on 15/May/17 Answered by mrW1 last updated on 16/May/17 $${x}>\mathrm{9} \\ $$$$\mathrm{log}\:\left(\mathrm{2}{x}−\mathrm{1}\right)\left({x}−\mathrm{9}\right)>\mathrm{2} \\ $$$$\left(\mathrm{2}{x}−\mathrm{1}\right)\left({x}−\mathrm{9}\right)>\mathrm{100} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{19}{x}−\mathrm{91}>\mathrm{0} \\…