Cần gấp ạ mong mn giúp e

Toán lớp 11 , bài 19 logarit

Đáp án:

Câu 3: $\log_32=\dfrac{a+b-1}{2a-b+1}$

Câu 4: $\dfrac{c+9ab}{3a\left(c+1\right)}$

Câu 5: $8.6$ độ Richter 

Giải thích các bước giải:

Câu 3:

Ta có:

$a=\log_518=\log_52\cdot 3^2=\log_52+2\log_53\to\log_52=a-2\log_53$

$b=\log_560=\log_55\cdot 2^2\cdot 3=1+2\log_52+\log_53$

$\to b=1+2(a-2\log_53)+\log_53$

$\to b=1+2a-3\log_53$

$\to 3\log_53=2a-b+1$

$\to\log_53=\dfrac{2a-b+1}3$

$\to \log_52=a-\dfrac{2a-b+1}3=\dfrac{a+b-1}3$

$\to \log_32=\dfrac{\log_52}{\log_53}=\dfrac{a+b-1}3:\dfrac{2a-b+1}3=\dfrac{a+b-1}{2a-b+1}$

Câu 4:

Ta có:

$a=\log_{27}5=\log_{3^3}5=\dfrac13\log_53\to \log_53=3a$

$b=\log_87=\log_{2^3}7=\dfrac13\log_27\to \log_27=3b$

$c=\log_23$

$\to \log_37=\dfrac{\log_27}{\log_23}=\dfrac{3b}c$

      $\log_52=\log_53\cdot\log_32=\dfrac{3a}c$

Ta có:

$\log_635$

$=\log_65+\log_67$

$=\dfrac1{\log_56}+\dfrac1{\log_76}$

$=\dfrac1{\log_52+\log_53}+\dfrac1{\log_72+\log_73}$

$=\dfrac1{\dfrac{3a}c+3a}+\dfrac1{\dfrac1{3b}+\dfrac{c}{3b}}$

$=\dfrac{c+9ab}{3a\left(c+1\right)}$

Câu 5:

Ta có:

$M=\log A-\log A_0=\log\dfrac{A}{A_0}$

Khi $M=8\to \log\dfrac{A}{A_0}=8$

Cường độ trận động đất ở Nam mỹ là:

$M'=\log\dfrac{4A}{A_0}=\log4+\log\dfrac{A}{A_0}=\log4+8\approx 8.6$ (độ Richter)