Question Number 64785 by Rio Michael last updated on 21/Jul/19 Answered by LPM last updated on 21/Jul/19 $$\left.\mathrm{1}\right)\:\mathrm{x}_{\mathrm{n}} \leqslant\:\mathrm{2}\:,\forall\:\mathrm{n}\geqslant\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{x}_{\mathrm{n}} \leqslant\mathrm{x}_{\mathrm{n}+\mathrm{1}} ,\:\forall\:\mathrm{n}\geqslant\mathrm{1} \\ $$$$\:\Rightarrow\:\mathrm{x}_{\mathrm{n}}…
Question Number 130244 by Adel last updated on 23/Jan/21 Commented by mr W last updated on 23/Jan/21 $${it}'{s}\:{not}\:{to}\:{understand}:\:{this}\:{question} \\ $$$${is}\:{asked}\:{already}\:{four}\:{times},\:{three} \\ $$$${times}\:{alone}\:{from}\:{you}!\:{what}'{s}\:{the} \\ $$$${sense}?\:{if}\:{you}\:{don}'{t}\:{understand}\:{the} \\…
Question Number 129451 by bemath last updated on 15/Jan/21 $$\:\mathrm{log}\:_{\mathrm{12}} \left(\sqrt{\mathrm{x}}\:+\:\sqrt[{\mathrm{4}}]{\mathrm{x}}\:\right)\:=\:\mathrm{log}\:_{\mathrm{9}} \left(\sqrt{\mathrm{x}}\:\right)\: \\ $$$$\:\mathrm{x}\:=\:? \\ $$ Answered by liberty last updated on 16/Jan/21 $$\:\mathrm{log}\:_{\mathrm{12}} \left(\sqrt{\mathrm{x}}\:+\:\sqrt[{\mathrm{4}}]{\mathrm{x}}\:\right)\:=\:\mathrm{log}\:_{\mathrm{3}}…
Question Number 129088 by Gulnoza last updated on 12/Jan/21 Answered by MJS_new last updated on 12/Jan/21 $$\mathrm{4}^{\mathrm{2log}_{\mathrm{4}} \:{x}} =\left(\mathrm{4}^{\mathrm{log}_{\mathrm{4}} \:{x}} \right)^{\mathrm{2}} ={x}^{\mathrm{2}} =\mathrm{25} \\ $$$$\Rightarrow\:{x}=\pm\mathrm{5}…
Question Number 128829 by bemath last updated on 10/Jan/21 $$\:\mathrm{If}\:\mathrm{x},\mathrm{y}\:\mathrm{and}\:\mathrm{x}\:\mathrm{in}\:\mathrm{HP}\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\mathrm{log}\:\left(\mathrm{x}+\mathrm{z}\right)\:+\mathrm{log}\:\left(\mathrm{x}+\mathrm{z}−\mathrm{y}\right)\:=\:\mathrm{2}\:\mathrm{log}\:\left(\mathrm{x}−\mathrm{z}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 128505 by Lian last updated on 08/Jan/21 $${simplify}:\:\:\:\frac{\mathrm{5}^{{log}_{\frac{\mathrm{4}}{\mathrm{5}}} \mathrm{5}} }{\mathrm{4}^{{log}_{\frac{\mathrm{4}}{\mathrm{5}}} \mathrm{4}} } \\ $$ Commented by liberty last updated on 08/Jan/21 $$=\:\mathrm{0}.\mathrm{05}\:=\:\frac{\mathrm{5}}{\mathrm{100}}=\frac{\mathrm{1}}{\mathrm{20}} \\…
Question Number 192918 by MATHEMATICSAM last updated on 31/May/23 Answered by aba last updated on 31/May/23 $$\mathrm{x}+\mathrm{1}=\mathrm{log}_{\mathrm{2a}} \left(\frac{\mathrm{bcd}}{\mathrm{2}}\right)+\mathrm{1}=\mathrm{log}_{\mathrm{2a}} \left(\frac{\mathrm{bcd}}{\mathrm{2}}\right)+\mathrm{log}_{\mathrm{2a}} \left(\mathrm{2a}\right)=\mathrm{log}_{\mathrm{2a}} \left(\mathrm{abcd}\right)\:\Rightarrow\frac{\mathrm{1}}{\mathrm{x}+\mathrm{1}}=\mathrm{log}_{\mathrm{abcd}} \left(\mathrm{2a}\right) \\ $$$$\mathrm{y}+\mathrm{1}=\mathrm{log}_{\mathrm{3b}} \left(\frac{\mathrm{acd}}{\mathrm{3}}\right)+\mathrm{1}=\mathrm{log}_{\mathrm{3b}}…
Question Number 192458 by Mingma last updated on 18/May/23 Answered by Frix last updated on 19/May/23 $$\mathrm{log}_{{x}+\frac{\mathrm{7}}{\mathrm{2}}} \:\left(\frac{{x}+\mathrm{7}}{\mathrm{2}{x}+\mathrm{3}}\right)^{\mathrm{2}} \:=\frac{\mathrm{ln}\:\left(\frac{{x}+\mathrm{7}}{\mathrm{2}{x}+\mathrm{3}}\right)^{\mathrm{2}} }{\mathrm{ln}\:\left({x}+\frac{\mathrm{7}}{\mathrm{2}}\right)}= \\ $$$$=\frac{\mathrm{2}\left(\mathrm{ln}\:\mid{x}+\mathrm{7}\mid\:−\mathrm{ln}\:\mid\mathrm{2}{x}+\mathrm{3}\mid\right)}{\mathrm{ln}\:\left(\mathrm{2}{x}+\mathrm{7}\right)\:−\mathrm{ln}\:\mathrm{2}}\geqslant\mathrm{0} \\ $$$$\frac{\mathrm{ln}\:\mid{x}+\mathrm{7}\mid\:−\mathrm{ln}\:\mid\mathrm{2}{x}+\mathrm{3}\mid}{\mathrm{ln}\:\left(\mathrm{2}{x}+\mathrm{7}\right)\:−\mathrm{ln}\:\mathrm{2}}\geqslant\mathrm{0} \\…
Question Number 192426 by mathlove last updated on 17/May/23 $${log}\left(−\mathrm{10}\right)\left(−\mathrm{10}\right)=? \\ $$ Answered by Frix last updated on 18/May/23 $$\mathrm{Unclear}: \\ $$$$\mathrm{log}\:\left({a}\right)\:\left({a}\right)\:=\left({a}\right)\mathrm{log}\:\left({a}\right) \\ $$$$−\mathrm{10}\:\mathrm{log}\:\left(−\mathrm{10}\right)\:=−\mathrm{10log}\:\left(\mathrm{10e}^{\mathrm{i}\pi} \right)…
Question Number 126389 by shaker last updated on 20/Dec/20 Answered by liberty last updated on 20/Dec/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\:\left[\frac{\mathrm{3cos}\:\mathrm{3}{x}}{\mathrm{sin}\:\mathrm{3}{x}}\right]}{\:\left[\frac{\mathrm{2cos}\:\mathrm{2}{x}}{\mathrm{sin}\:\mathrm{2}{x}}\right]}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{3cos}\:\mathrm{3}{x}}{\mathrm{sin}\:\mathrm{3}{x}}×\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2cos}\:\mathrm{2}{x}} \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{2}}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{sin}\:\mathrm{3}{x}}\:=\:\mathrm{1} \\ $$ Answered…