Câu hỏi của Như Quỳnh

Question

1. cho a=log3 2 và b=log3 5. tính các logarit sau theo a, b; A=log3 80, B=log3 37,5

2. cho log10 3=a, log5=b. tính C=log30 8 theo a, b

3. cho log27 5=a, log8 7=b, log2 3=c. tính D log6 35 theo a, b,...

Akai Haruma · Accepted Answer

Bài 1:

$A=\log_380=\log_3(2^4.5)=\log_3(2^4)+\log_3(5)$

$=4\log_32+\log_35=4a+b$

$B=\log_3(37,5)=\log_3(2^{-1}.75)=\log_3(2^{-1}.3.5^2)$

$=\log_3(2^{-1})+\log_33+\log_3(5^2)=-\log_32+1+2\log_35$

$=-a+1+2b$Câu trả lời: Bài 1:

$A=\log_380=\log_3(2^4.5)=\log_3(2^4)+\log_3(5)$

$=4\log_32+\log_35=4a+b$

$B=\log_3(37,5)=\log_3(2^{-1}.75)=\log_3(2^{-1}.3.5^2)$

$=\log_3(2^{-1})+\log_33+\log_3(5^2)=-\log_32+1+2\log_35$

$=-a+1+2b$

Akai Haruma · Answer

Bài 2:

$\log_{30}8=\frac{\log 8}{\log 30}=\frac{\log (2^3)}{\log (10.3)}=\frac{3\log2}{\log 10+\log 3}$

$=\frac{3\log (\frac{10}{5})}{1+\log 3}=\frac{3(\log 10-\log 5)}{1+\log 3}=\frac{3(1-b)}{1+a}$Câu trả lời: Bài 2:

$\log_{30}8=\frac{\log 8}{\log 30}=\frac{\log (2^3)}{\log (10.3)}=\frac{3\log2}{\log 10+\log 3}$

$=\frac{3\log (\frac{10}{5})}{1+\log 3}=\frac{3(\log 10-\log 5)}{1+\log 3}=\frac{3(1-b)}{1+a}$