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Category: Logarithms

Question-126389

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…

Question-191392

Question Number 191392 by Mingma last updated on 23/Apr/23 Answered by Frix last updated on 23/Apr/23 $$\frac{\sqrt{\mathrm{ln}\:{n}}}{\:\sqrt{\mathrm{ln}\:{b}}}=\frac{\mathrm{ln}\:{n}}{\mathrm{2ln}\:{b}}\:\Rightarrow\:\mathrm{ln}\:{n}\:=\mathrm{4ln}\:{b} \\ $$$$\frac{{b}\mathrm{ln}\:{n}}{\mathrm{ln}\:{b}}=\frac{\mathrm{ln}\:{bn}}{\mathrm{ln}\:{b}}\:\Rightarrow\:\mathrm{ln}\:{n}\:=\frac{\mathrm{ln}\:{b}}{{b}−\mathrm{1}} \\ $$$$\Rightarrow \\ $$$$\mathrm{4ln}\:{b}\:=\frac{\mathrm{ln}\:{b}}{{b}−\mathrm{1}}\:\Rightarrow\:{b}=\frac{\mathrm{5}}{\mathrm{4}}\:\Rightarrow\:{n}=\frac{\mathrm{625}}{\mathrm{256}} \\ $$…

Question-191155

Question Number 191155 by mathlove last updated on 19/Apr/23 Answered by mahdipoor last updated on 19/Apr/23 $${log}\left({xy}\right)={log}\left({x}\right)+{log}\left({y}\right)\Rightarrow{x}={a}\:,\:{y}=\mathrm{1}+\frac{{b}}{{a}}\:\Rightarrow \\ $$$${log}\left({a}+{b}\right)={log}\left({a}\right)+{log}\left(\mathrm{1}+{b}/{a}\right) \\ $$$$…….. \\ $$$${b}=\mathrm{10}^{{B}} \:,\:{a}=\mathrm{10}^{{A}} \:,\:{c}=\mathrm{10}^{{C}}…

5log-4-2-3-6-6log-8-3-2-

Question Number 59401 by Pranay last updated on 09/May/19 $$\mathrm{5}{log}_{\mathrm{4}\sqrt{\mathrm{2}}} \left(\mathrm{3}−\sqrt{\mathrm{6}}\:\right)\:−\mathrm{6}{log}_{\mathrm{8}} \left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right) \\ $$ Answered by prakash jain last updated on 10/May/19 $$\mathrm{4}\sqrt{\mathrm{2}}=\left(\sqrt{\mathrm{2}}\right)^{\mathrm{5}} \\ $$$$\mathrm{8}=\left(\sqrt{\mathrm{2}}\right)^{\mathrm{6}}…