Cho a = log(30)3, b= log(30)5

Giải Toán 12 Bài 3: Lôgarit

Bài 5 trang 68 Toán lớp 12 Giải tích:

a) Cho $a={\log }_{30}3,b={\log }_{30}5$. Hãy tính ${\log }_{30}1350$ theo a, b.

b) Cho $c={\log }_{15}3$. Hãy tính ${\log }_{25}15$ theo c.

Lời giải:

a) Ta có:

${\log }_{30}1350={\log }_{30}\left({30.3}^2.5\right)$

$={\log }_{30}30+{\log }_{30}{3}^2+{\log }_{30}5$

$=1+2{\log }_{30}3+{\log }_{30}5$

 = 1 + 2a + b

Vậy ${\log }_{30}1350$ = 1 + 2a + b.

b) Cách 1: Ta có

${\log }_{25}15=\frac{1}{{\log }_{15}25}=\frac{1}{{\log }_{15}{5}^2}$

$=\frac{1}{2{\log }_{15}5}=\frac{1}{2{\log }_{15}\left(15:3\right)}$

$=\frac{1}{2\left({\log }_{15}15-{\log }_{15}3\right)}$

$=\frac{1}{2\left(1-{\log }_{15}3\right)}=\frac{1}{2\left(1-c\right)}$

Cách 2:

${\log }_{25}15={\log }_{5^2}15=\frac{1}{2}{\log }_{5}15=\frac{1}{2}{\log }_5\left(5.3\right)=\frac{1}{2}\left({\log }_{5}5+{\log }_{5}3\right)=\frac{1}{2}\left(1+{\log }_{5}3\right)$

Lại có: c = ${\log }_{15}3$

$\Rightarrow \frac{1}{c}=\frac{1}{{\log }_{15}3}={\log }_{3}15={\log }_3\left(3.5\right)=lo{g}_{3}3+{\log }_{3}5=1+{\log }_{3}5\Rightarrow {\log }_{3}5=\frac{1}{c}-1=\frac{1-c}{c}\Rightarrow {\log }_{5}3=\frac{c}{1-c}$

Khi đó: 

${\log }_{25}15=\frac{1}{2}\left(1+\frac{c}{1-c}\right)=\frac{1}{2}.\frac{1-c+c}{1-c}=\frac{1}{2\left(1-c\right)}$