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 ${e^{\ln (x)}} = x$