Category: Logarithms

Question Number 128505 by Lian last updated on 08/Jan/21 $$simplify:\frac{5^{{log}_{\frac{4}{5}}5}}{4^{{log}_{\frac{4}{5}}4}}$$ 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 192458 by Mingma last updated on 18/May/23 Answered by Frix last updated on 19/May/23 $${\log }_{x+\frac{7}{2}}{\left(\frac{x+7}{2x+3}\right)}^2=\frac{\ln {\left(\frac{x+7}{2x+3}\right)}^2}{\ln (x+\frac{7}{2})}=$$$$=\frac{2(\ln \mid x+7\mid -\ln \mid 2x+3\mid )}{\ln (2x+7)-\ln 2}⩾0$$$$\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 126389 by shaker last updated on 20/Dec/20 Answered by liberty last updated on 20/Dec/20 $$\underset{x\to 0}{\lim }\frac{\left[\frac{3\cos 3x}{\sin 3x}\right]}{\left[\frac{2\cos 2x}{\sin 2x}\right]}=\underset{x\to 0}{\lim }\frac{3\cos 3x}{\sin 3x}\times \frac{\sin 2x}{2\cos 2x}$$$$=\frac{3}{2}\times \underset{x\to 0}{\lim }\frac{\sin 2x}{\sin 3x}=1$$ Answered…

Question Number 191516 by leandrosriv02 last updated on 25/Apr/23 Terms of Service Privacy Policy Contact: info@tinkutara.com

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