Respuesta :
Answer: The required value of the given expression is - 2.
Step-by-step explanation: We are given to find the value of the following logarithmic expression:
[tex]E=\log_6\dfrac{1}{36}.[/tex]
We will be using the following logarithmic properties:
[tex](i)~\log_ab=\dfrac{\log b}{\log a},\\\\(ii)\log a^b=b\log a.[/tex]
We have
[tex]E\\\\=\log_6\dfrac{1}{36}\\\\\\=\log_6\dfrac{1}{6^2}\\\\\\=\log_66^{-2}\\\\\\=\dfrac{\log6^{-2}}{\log 6}\\\\\\=\dfrac{-2\log6}{\log6}\\\\\\=-2.[/tex]
Thus, the required value of the given expression is - 2.
The value of [tex]{\log _6}\left( {\dfrac{1}{{36}}} \right)[/tex] is [tex]\boxed{{{\log }_6}\left( {\frac{1}{{36}}} \right) = - 2}.[/tex]
Further explanation:
The logarithm properties can be expressed as follows,
(a). [tex]{\log _a}a = 1[/tex]
(b). [tex]{\log _a}{a^b} = b[/tex]
(c). [tex]{\log _a}b = \dfrac{{\log b}}{{\log a}}[/tex]
Given:
The expression is [tex]{\log _6}\left( {\dfrac{1}{{36}}} \right).[/tex]
Explanation:
The given expression is [tex]{\log _6}\left( {\dfrac{1}{{36}}} \right).[/tex]
Consider the expression as A.
[tex]\begin{aligned}{\text{A}} &= {\log _6}\left( {\frac{1}{{36}}} \right)\\&= {\log _6}\left( {\frac{1}{{{6^2}}}} \right)\\&= {\log _6}\left( {{6^{ - 2}}} \right)\\&= - 2\\\end{aligned}[/tex]
The value of [tex]{\log _6}\left( {\dfrac{1}{{36}}} \right)[/tex] is [tex]\boxed{{{\log }_6}\left( {\frac{1}{{36}}} \right) = - 2}.[/tex]
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Answer details:
Grade: High school
Subject: Mathematics
Chapter: Number system
Keywords: place value, base value, decimal expansion, value, log value, log6, 1/36.
