ĐGTD ĐH Bách khoa - Tư duy Toán học - Hàm số có Logarit

a < 0, b > 0

$0<a\ne \mathrm{1,}b<0$

$0<a\ne \mathrm{1,}b>0$

$0<a\ne \mathrm{1,0}<b\ne 1$

$2{\log }_a\sqrt{b}={\log }_{a}b$

${\log }_a\sqrt{b}=2{\log }_{a}b$

${\log }_a\sqrt[3]{b}=\frac{1}{3}$

${\log }_a\sqrt[3]{b}=-\frac{1}{3}{\log }_{a}b$

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

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

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

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

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

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

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

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

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

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

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

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

0 < a < 1

a > 3

a > 1

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

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

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

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

${\log }_{2}16={\log }_{3}81$

${\log }_{3}9=3$

${\log }_{4}16={\log }_{2}8$

$2^{{\log }_{2}3}=5^{{\log }_{3}5}$

$2^{{\log }_{2}3}=5^{{\log }_{5}3}$

$5^{{\log }_{5}3}={\log }_{2}3$

$2^{{\log }_{2}4}=2$

2

-8

-2

$\frac{1}{2}$

$P={\log }_2{\left(\frac{a}{b}\right)}^2$

$P={\log }_2\left(\frac{2a}{b^2}\right)$

$P={\log }_2\left(2a{b}^2\right)$

$P={\log }_2{\left(ab\right)}^2$

$P=\frac{\sqrt{3}}{3}$

$P=\frac{1}{3}$

P = 27

$P=3\sqrt{3}$

$P=\frac{2a+b+3}{a}$

$P=\frac{a+b+4}{a}$

$P=\frac{a+b+3}{a}$

$P=\frac{a+2b+3}{a}$

${\log }_{6}45=\frac{2a^2-2ab}{ab}$

${\log }_{6}45=\frac{2a^2-2ab}{ab+b}$

${\log }_{6}45=\frac{a+2ab}{ab+b}$

${\log }_{6}45=\frac{a+2ab}{ab}$

a > b > c

c > a > b

c > b > a

b > a > c

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

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

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

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

$\frac{1-a}{a-2}$

$\frac{2a-1}{a-2}$

$\frac{a-1}{2a-2}$

$\frac{1-2a}{a-2}$

$\frac{10}{a-1}$

$\frac{2}{5(a-1)}$

$\frac{5}{2a-2}$

$\frac{5}{2a+1}$

$P=\frac{ab+2a+2}{b}$

$P=\frac{ab-a+2}{b}$

$P=\frac{ab+a-2}{b}$

$P=\frac{ab-a-2}{b}$

${\log }_{3}90=\frac{a-2b+1}{b+1}$

${\log }_{3}90=\frac{a+2b-1}{b-1}$

${\log }_{3}90=\frac{2a-b+1}{a+1}$

${\log }_{3}90=\frac{2a+b-1}{a-1}$

$a^2{p}^4$

4p + 2

4p + 2a

$p^4+2a$

${\log }_{12}80=\frac{2a^2-2ab}{ab+b}$

${\log }_{12}80=\frac{a+2ab}{ab}$

${\log }_{12}80=\frac{a+2ab}{ab+b}$

${\log }_{12}80=\frac{2a^2-2ab}{ab}$

${\log }_{2}7=\frac{a}{1-b}$

${\log }_{2}7=\frac{b}{1-a}$

${\log }_{2}7=\frac{a}{1+b}$

${\log }_{2}7=\frac{b}{1+a}$

$a=\frac{1}{b}$

a = b

$a=\frac{1}{b^2}$

$a=b^2$

18

20

19

21

$2\log \left(a+2b\right)=5\left(\log a+\log b\right)$

$\log \left(a+1\right)+\log b=1$

$\log \frac{a+2b}{3}=\frac{\log a+\log b}{2}$

$5\log \left(a+2b\right)=\log a-\log b$

T = -3

T = 3

T = -1

T = 1

x < y

x > y

$x\le y$

$x\ge y$

4

2

3

1

T = 7

T = 12

T = 13

T = 21

$\frac{2ab}{1+b}$

$\frac{ab}{1+b}$

$\frac{a}{1+b}$

$\frac{b}{1+b}$

900

1350

1050

1200

8

10

9

90

$t\approx \mathrm{1,65}$giờ

$t\approx \mathrm{1,61}$giờ

$t\approx \mathrm{1,63}$giờ

$t\approx \mathrm{1,50}$giờ

$\ln \left(a+2b\right)-2\ln 2=\ln a+\ln b$

$\ln \left(a+2b\right)=\frac{1}{2}(\ln a+\ln b)$

$\ln \left(a+2b\right)-2\ln 2=\frac{1}{2}(\ln a+\ln b)$

$\ln \left(a+2b\right)+2\ln 2=\frac{1}{2}(\ln a+\ln b)$

P = 6

$P=3+2\sqrt{2}$

$P=2+3\sqrt{2}$

$P=\sqrt{3}+\sqrt{17}$

$a^3+b^3=8a^{2}b-a{b}^2$

$a^3+b^3=3\left(8a^{2}b+a{b}^2\right)$

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

$a^3+b^3=3\left(8a^{2}b-a{b}^2\right)$

$A=T{e}^{2r}$

$A=T{e}^{-2r}$

$A=\frac{T}{e^{-2r}}$

$A=2T{e}^{-r}$