Hello there. To solve this question, we'll have to remember some properties about changing bases in logarithms.
Given the logarithm:
[tex]\log_3\left(\dfrac{1}{4}\right)[/tex]
We first use the properties:
[tex]\begin{gathered} \log_c\left(\dfrac{a}{b}\right)=\log_c(a)-\log_c(b)\text{ and } \\ \\ \log_c(1)=0,0In this case, we get:
[tex]\begin{gathered} \log_3\left(\dfrac{1}{4}\right)=\log_3(1)-\log_3(4)=0-\log_3(4) \\ \\ \Rightarrow-\log_3(4) \end{gathered}[/tex]
Now, we use the change of basis formula:
[tex]\log_c(a)=\dfrac{\log_d(a)}{\log_d(c)}[/tex]
When changing for a basis d greater than zero and not equal to 1.
With this, we have that
[tex]-\log_3(4)=-\dfrac{\log_{10}(4)}{\log_{10}(3)}[/tex]
Applying the property:
[tex]\log_c(a^b)=b\cdot\log_c(a)[/tex]
and knowing that 4 = 2², we get
[tex]-\dfrac{2\log_{10}(2)}{\log_{10}(3)}[/tex]
We chose base 10 log because we know the following values for:
[tex]\begin{gathered} \log_{10}(2)\approx0.3010 \\ \\ \log_{10}(3)\approx0.477 \end{gathered}[/tex]
Hence the approximation for what we want is
[tex]\log_3\left(\dfrac{1}{4}\right)\approx-\dfrac{2\cdot0.3010}{0.477}=-1.262[/tex]
This is the answer we're looking for.
