Question

How do you evaluate \[{\log _3}\left( {\dfrac{1}{9}} \right)\] ?

Answer 1

Hint: We can solve the given problem if we know the basic logarithmic rules. In the given problem we have logarithm to the base 3. So we need to apply a change of base formula to it. That is \[{\log _b}a = \dfrac{{{{\log }_{10}}a}}{{{{\log }_{10}}b}}\] . We also know that the power rule for logarithm, that is \[{\log _b}({m^n}) = n{\log _b}(m)\] . Also we have quotient rule for logarithm that is
 \[\log \left( {\dfrac{m}{n}} \right) = \log m - \log n\] . Using these properties we can solve the given problem.

Complete step-by-step answer:
Given , \[{\log _3}\left( {\dfrac{1}{9}} \right)\] .
We know for change of base in logarithm,
 \[{\log _b}a = \dfrac{{{{\log }_{10}}a}}{{{{\log }_{10}}b}}\] .
Comparing with given problem we have
 \[a = \dfrac{1}{9}\] and \[b = 3\] .
Then we have,
 \[ \Rightarrow {\log _3}\left( {\dfrac{1}{9}} \right) = \dfrac{{{{\log }_{10}}\left( {\dfrac{1}{9}} \right)}}{{{{\log }_{10}}3}}\]
 \[ = \dfrac{{\log \left( {\dfrac{1}{9}} \right)}}{{\log 3}}\]
We know quotient rule for logarithm that is
 \[\log \left( {\dfrac{m}{n}} \right) = \log m - \log n\]
Applying this in the numerator term we have
 \[ = \dfrac{{\log \left( 1 \right) - \log (9)}}{{\log 3}}\]
Splitting the terms we have,
 \[ = \dfrac{{\log \left( 1 \right)}}{{\log 3}} - \dfrac{{\log (9)}}{{\log 3}}\]
We can write 9 as \[{3^2}\]
 \[ = \dfrac{{\log \left( 1 \right)}}{{\log 3}} - \dfrac{{\log ({3^2})}}{{\log 3}}\]
We know the power rule for logarithm, that is \[{\log _b}({m^n}) = n{\log _b}(m)\] .
 \[ = \dfrac{{\log \left( 1 \right)}}{{\log 3}} - \dfrac{{2\log (3)}}{{\log 3}}\]
Cancelling we have
 \[ = \dfrac{{\log \left( 1 \right)}}{{\log 3}} - 2\]
We know \[\log \left( 1 \right) = 0\]
 \[ = - 2\] .
Thus, we have
 \[ \Rightarrow {\log _3}\left( {\dfrac{1}{9}} \right) = - 2\]
So, the correct answer is “-2”.

Note: We know the product rule for logarithms can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms. That is \[{\log _b}(mn) = {\log _b}(m) + {\log _b}(n)\] . Also we have common and natural logarithmic functions. Common logarithmic function is a logarithm with base 10 is common logarithm. A natural logarithm is different. When the base of the common logarithm is 10, the base of a natural logarithm is number \[e\] . Although \[e\] represents a variable it is a fixed irrational number that is equal to 2.71828… Logarithm and \[e\] cancel out. for example if we have \[{e^{\ln (x)}}\] then the result is ‘x’. That is \[{e^{\ln (x)}} = x\]