Given:
[tex]\log_34\approx 1.262[/tex]
[tex]\log_37\approx 1.771[/tex]
To find:
The value of [tex]\log_3\left(\dfrac{4}{49}\right)[/tex].
Solution:
We have,
[tex]\log_34\approx 1.262[/tex]
[tex]\log_37\approx 1.771[/tex]
Using properties of log, we get
[tex]\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_349[/tex] [tex]\left[\because \log_a\dfrac{m}{n}=\log_am-\log_an\right][/tex]
[tex]\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_37^2[/tex]
[tex]\log_3\left(\dfrac{4}{49}\right)=\log_34-2\log_37[/tex] [tex][\log x^n=n\log x][/tex]
Substitute [tex]\log_34\approx 1.262[/tex] and [tex]\log_37\approx 1.771[/tex].
[tex]\log_3\left(\dfrac{4}{49}\right)=1.262-2(1.771)[/tex]
[tex]\log_3\left(\dfrac{4}{49}\right)=1.262-3.542[/tex]
[tex]\log_3\left(\dfrac{4}{49}\right)=-2.28[/tex]
Therefore, the value of [tex]\log_3\left(\dfrac{4}{49}\right)[/tex] is [tex]-2.28[/tex].
