20 câu Trắc nghiệm Logarit có đáp án (Nhận biết)

${\log }_{5}2$

2log5

log(25)

${\log }_{2}5$

a<0,b>0

$0<a\ne 1,b<0$

$0<a\ne 1,b>0$

$0<a\ne 1,0<b\ne 1$

$x\le 3$

x>3

$x\ge 3$

x<3

$a^b=N$

${\log }_{a}N=b$

$a^N=b$

$b^N=a$

$x=\frac{1}{8}$

$x=\sqrt{3}$

$x=\frac{1}{6}$

$x=\frac{3}{8}$

${\log }_{a}1=1$

${\log }_{a}a=a$

${\log }_{a}1=a$

${\log }_{a}a=1$

${\log }_{2}1=1$

${\log }_{2}2=2$

${\log }_{2}1=2$

${\log }_{2}2=1$

${\log }_a{a}^b=b$

${\log }_a{a}^b=a^b$

$a^{{\log }_{a}b}=b$

$a^{{\log }_{a}b}={\log }_a{a}^b$

${\log }_a\left(bc\right)={\log }_{a}b+{\log }_{b}c$

${\log }_a\frac{b}{c}={\log }_{a}b+{\log }_{a}c$

${\log }_a\frac{b}{c}=\frac{{\log }_{a}b}{{\log }_{a}c}$

${\log }_a\left(bc\right)={\log }_{a}b+{\log }_{a}c$

$\frac{{\log }_{a}b}{{\log }_{a}c}={\log }_{b}c$

$\frac{{\log }_{a}b}{{\log }_{a}c}={\log }_a\left(\frac{b}{c}\right)$

$\frac{{\log }_{a}b}{{\log }_{a}c}={\log }_{c}b$

$\frac{{\log }_{a}b}{{\log }_{a}c}={\log }_a\left(b-c\right)$

${\log }_a{b}^n=n{\log }_{a}b$

${\log }_a\sqrt[n]{b}=\frac{1}{n}{\log }_{a}b$

${\log }_a\frac{1}{b}=-{\log }_{a}b$

${\log }_a\sqrt[n]{b}=-n{\log }_{a}b$

${\log }_a\frac{x}{y}={\log }_{a}x+{\log }_{a}y$

${\log }_a\frac{x}{y}={\log }_a\left(x-y\right)$

${\log }_a\frac{x}{y}={\log }_{a}x-{\log }_{a}y$

${\log }_a\frac{x}{y}=\frac{{\log }_{a}x}{{\log }_{a}y}$

${\log }_{a^n}b=-n{\log }_{a}b$

${\log }_{a^n}b=\frac{1}{n}{\log }_{a}b$

${\log }_{a^n}b=-\frac{1}{n}{\log }_{a}b$

${\log }_{a^n}b=n{\log }_{a}b$

${\log }_{a}b.{\log }_{b}c={\log }_{a}c$

${\log }_{b}c=\frac{{\log }_{a}b}{{\log }_{a}c}$

${\log }_{a}b={\log }_{c}b-{\log }_{c}a$

${\log }_{a}b+{\log }_{b}c={\log }_{a}c$

${\log }_{a}b>{\log }_{a}c$

${\log }_{a}b<{\log }_{a}c$

${\log }_{a}b<{\log }_{b}c$

${\log }_{a}b>{\log }_{c}b$

${\log }_{a}b=0$

${\log }_{a}b>0$

${\log }_{a}b<0$

${\log }_{a}b=1$

$\ln a=b\iff b={10}^a$

$\ln a=b\iff a=e^b$

$\ln a=b\iff b=e^a$

$\ln a=b\iff a={10}^b$

x=e

$x={\left(-1\right)}^e$

$x=\frac{1}{e}$

$x=\sqrt{e}$

ln(ab)=lna+lnb

$\ln {\left(a^2-b\right)}^3=3\ln \left(a^2-b\right)$

$\ln \left(\frac{a}{b}\right)=\ln \left(a\right)-\ln (-b)$

$\ln {\left(\frac{a}{b}\right)}^2=\ln a^2-\ln b^2$