Question

How do you evaluate ${\log _6}\left( {\dfrac{1}{{216}}} \right)$?

Answer 1

The above question is based on the concept of logarithms. The main approach towards solving the expression is by applying the logarithm properties. On applying the properties of logarithm and further on simplifying we will get a value for the above expression.

Complete step by step solution:
Logarithm is the exponent or power to which base must be raised to yield a given number. When expressed mathematically x is the logarithm of base n to the base b if ${b^x} = n$,then we can write it has
$x = {\log _b}n$
The simplest way to understand the logarithm is given for example-Number of 2s we multiply to get 8.
$2 \times 2 \times 2 = 8$. So, we have to multiply 3 of the 2s to get 8.
Therefore, the number of 2s we need to multiply to get 8 is 3
${\log _2}\left( 8 \right) = 3$
So now apply it on the expression.
Rewrite it as an equation by equating with x.
${\log _6}\left( {\dfrac{1}{{216}}} \right) = x$
216 is the perfect cube of the number 6. So, we can write in the form of 6.
${\log _6}\left( {\dfrac{1}{{{{\left( 6 \right)}^3}}}} \right) = x$
If x and b are positive real numbers and b does not equal 1, then ${\log _b}x = y$ is equivalent to
${b^y} = x$

$${6^x} = \dfrac{1}{{{{\left( 6 \right)}^3}}} \\\ {6^x} = {6^{ - 3}} \\\$$

Since the bases are the same, the two expressions are only equal if the exponents are also equal. Therefore, x=-3.

Note: An important thing to note is that the term ${\log _6}\left( {\dfrac{1}{{216}}} \right)$ can be explained in a way that $\dfrac{1}{6}$ is multiplied 3 times to get $\dfrac{1}{{216}}$. So 6 when taken in the numerator the power becomes -3. Thus, the value we get is -3.