Answer :
To solve the expression \(\log_6 \frac{1}{36}\), we need to determine the power to which the base 6 must be raised to obtain \(\frac{1}{36}\).
Firstly, let's consider \(\frac{1}{36}\) in terms of \(6\):
We know that:
[tex]\[ 36 = 6^2 \][/tex]
Therefore, \(\frac{1}{36}\) can be written as:
[tex]\[ \frac{1}{36} = \frac{1}{6^2} = 6^{-2} \][/tex]
Now, we need to find the logarithm:
[tex]\[ \log_6(6^{-2}) \][/tex]
The logarithm \(\log_b(a)\) of a number \(a\) with base \(b\) is the exponent \(x\) such that \(b^x = a\).
In our case:
[tex]\[ \log_6(6^{-2}) = -2 \][/tex]
Thus, the value of \(\log_6 \frac{1}{36}\) is \(-2\).
So, the correct answer is:
[tex]\[\boxed{-2}\][/tex]
Firstly, let's consider \(\frac{1}{36}\) in terms of \(6\):
We know that:
[tex]\[ 36 = 6^2 \][/tex]
Therefore, \(\frac{1}{36}\) can be written as:
[tex]\[ \frac{1}{36} = \frac{1}{6^2} = 6^{-2} \][/tex]
Now, we need to find the logarithm:
[tex]\[ \log_6(6^{-2}) \][/tex]
The logarithm \(\log_b(a)\) of a number \(a\) with base \(b\) is the exponent \(x\) such that \(b^x = a\).
In our case:
[tex]\[ \log_6(6^{-2}) = -2 \][/tex]
Thus, the value of \(\log_6 \frac{1}{36}\) is \(-2\).
So, the correct answer is:
[tex]\[\boxed{-2}\][/tex]
