Question

How do you evaluate ${\log _{\dfrac{1}{{11}}}}\left( {\dfrac{1}{{121}}} \right)$?

Answer 1

Hint: Here we will use the logarithmic properties to simplify the given expression. We will use division rule: ${\log _a}\left( {\dfrac{x}{y}} \right) = {\log _a}x - {\log _a}y$also apply the tangent and cot identity and then will simplify for the resultant value.

Complete step-by-step solution:
Take the given expression: ${\log _{\dfrac{1}{{11}}}}\left( {\dfrac{1}{{121}}} \right)$
Apply the change of base property: ${\log _a}b = \dfrac{{\log b}}{{\log a}}$
${\Rightarrow \log _{\dfrac{1}{{11}}}}\left( {\dfrac{1}{{121}}} \right) = \dfrac{{\log \left( {\dfrac{1}{{121}}} \right)}}{{\log \left( {\dfrac{1}{{11}}} \right)}}$
Now, Apply the property: ${\log _a}\left( {\dfrac{x}{y}} \right) = {\log _a}x - {\log _a}y$
${\Rightarrow \log _{\dfrac{1}{{11}}}}\left( {\dfrac{1}{{121}}} \right) = \dfrac{{\log 1 - \log 121}}{{\log 1 - \log 11}}$
Now, we know that $\log 1 = 0$apply in the above expression –
$\Rightarrow {\log _{\dfrac{1}{{11}}}}\left( {\dfrac{1}{{121}}} \right) = \dfrac{{0 - \log 121}}{{0 - \log 11}}$
Simplify the above expression –
$\Rightarrow {\log _{\dfrac{1}{{11}}}}\left( {\dfrac{1}{{121}}} \right) = \dfrac{{ - \log 121}}{{ - \log 11}}$
Negative sign from the numerator and the denominator cancels each other.
$\Rightarrow {\log _{\dfrac{1}{{11}}}}\left( {\dfrac{1}{{121}}} \right) = \dfrac{{\log 121}}{{\log 11}}$
Now, apply the squares of the number property –
$\Rightarrow {\log _{\dfrac{1}{{11}}}}\left( {\dfrac{1}{{121}}} \right) = \dfrac{{\log {{11}^2}}}{{\log 11}}$
By property $\log {a^b} = b\log a$so apply in the above expression –
$\Rightarrow {\log _{\dfrac{1}{{11}}}}\left( {\dfrac{1}{{121}}} \right) = \dfrac{{2\log 11}}{{\log 11}}$
Like terms from the numerator and the denominator cancels each other. Therefore, remove the numerator and the denominator of the above expression.
$\Rightarrow {\log _{\dfrac{1}{{11}}}}\left( {\dfrac{1}{{121}}} \right) = 2$
This is the required solution.

Thus the required solution is ${\log _{\dfrac{1}{{11}}}}\left( {\dfrac{1}{{121}}} \right) = 2$.

Additional Information: Also refer to the below properties and rules of the logarithm.
Product rule: ${\log _a}xy = {\log _a}x + {\log _a}y$
Quotient rule: ${\log _a}\dfrac{x}{y} = {\log _a}x - {\log _a}y$
Power rule: ${\log _a}{x^n} = n{\log _a}x$
 Base rule:${\log _a}a = 1$
Change of base rule: ${\log _a}M = \dfrac{{\log M}}{{\log N}}$
Know the difference between ln and log and apply its properties accordingly. Logarithms are the ways to figure out which exponents we need to multiply into the specific number. Log is defined for the base $10$ and ln is denoted for the base e. “e” is an irrational and transcendental number which can be expressed as $e = 2.71828$. You can convert ln to log by using the relation such as
$\ln (x) = \log x \div \log (2.71828)$

Note: In other words, the logarithm is the power to which the number must be raised in order to get some other. Always remember the standard properties of the logarithm.... Product rule, quotient rule and the power rule. The basic logarithm properties are most important and the solution solely depends on it, so remember and understand its application properly. Be good in multiples and know the concepts of square and square root and apply accordingly.