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…
Question Number 126217 by aristarque last updated on 18/Dec/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 191516 by leandrosriv02 last updated on 25/Apr/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 191518 by leandrosriv02 last updated on 25/Apr/23 Commented by MATHEMATICSAM last updated on 26/Apr/23 $$\mathrm{No}\:\mathrm{need}\:\mathrm{to}\:\mathrm{use}\:\mathrm{log} \\ $$$${a}^{{m}} \:=\:{a}^{{n}} \:\mathrm{then}\:{m}\:=\:{n} \\ $$ Terms of…
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 Number 191155 by mathlove last updated on 19/Apr/23 Answered by mahdipoor last updated on 19/Apr/23 $${log}\left({xy}\right)={log}\left({x}\right)+{log}\left({y}\right)\Rightarrow{x}={a}\:,\:{y}=\mathrm{1}+\frac{{b}}{{a}}\:\Rightarrow \\ $$$${log}\left({a}+{b}\right)={log}\left({a}\right)+{log}\left(\mathrm{1}+{b}/{a}\right) \\ $$$$…….. \\ $$$${b}=\mathrm{10}^{{B}} \:,\:{a}=\mathrm{10}^{{A}} \:,\:{c}=\mathrm{10}^{{C}}…
Question Number 59401 by Pranay last updated on 09/May/19 $$\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}} \\ $$$$\mathrm{8}=\left(\sqrt{\mathrm{2}}\right)^{\mathrm{6}}…
Question Number 124880 by WilliamsErinfolami last updated on 06/Dec/20 $${Solve}\:{for}\:{x} \\ $$$${x}^{{log}_{\mathrm{5}} \mathrm{3}{x}} =\mathrm{12}{x} \\ $$ Answered by mr W last updated on 06/Dec/20 $$\left(\mathrm{log}_{\mathrm{5}}…