Question

Find the logarithms of 0⋅0001 to base 0.001, and 0.1 to base $9\sqrt 3 $.

Answer 1

Hint: Use the change of base property of logarithm to change the given logs into common log. Then try to use other properties of log to simplify and evaluate the values. You may have to refer to the common log table.

Complete step by step answer:
Given in the problem we need to find the logarithms of 0⋅0001 to base 0.001, and 0.1 to base $9\sqrt 3 $.
In mathematics, the logarithm is the inverse function to exponentiation. That means the logarithm of a given number x is the exponent to which another fixed number, the base b, must be raised, to produce that number x.
The logarithm of $x$ to the base $b$is written as ${\log _b}x$.
Given in the problem, we need to find the values of
${\log _{0.001}}\left( {0.0001} \right)$
${\log _{9\sqrt 3 }}\left( {0.1} \right)$
Solving the first one.
We need to find the value of ${\log _{0.001}}\left( {0.0001} \right)$.
One of the properties of log is ${\log _b}b = 1$. Hence, we need to transform (i) in this form in order to evaluate the value.
The change of base property of log is defined as:
${\log _b}x = \dfrac{{{{\log }_a}x}}{{{{\log }_a}b}}{\text{ (1)}}$
Using this property in (i) with $a = 10$, we get
$
  {\log _{0.001}}\left( {0.0001} \right) = \dfrac{{{{\log }_{10}}0.0001}}{{{{\log }_{10}}0.001}} \\
   \Rightarrow {\log _{0.001}}\left( {0.0001} \right) = \dfrac{{{{\log }_{10}}{{10}^{ - 4}}}}{{{{\log }_{10}}{{10}^{ - 3}}}} \\
$
Using the property, ${\log _b}{x^c} = c{\log _b}x$in above, we get
$ \Rightarrow {\log _{0.001}}\left( {0.0001} \right) = \dfrac{{ - 4{{\log }_{10}}10}}{{ - 3{{\log }_{10}}10}}$
Since as stated above, ${\log _b}b = 1$, using this in above with $b = 10$, we get,
$ \Rightarrow {\log _{0.001}}\left( {0.0001} \right) = \dfrac{{ - 4.1}}{{ - 3.1}} = \dfrac{4}{3}$
Therefore, the value of ${\log _{0.001}}\left( {0.0001} \right)$ is $\dfrac{4}{3}$.
Similarly solving the second part.
We have, ${\log _{9\sqrt 3 }}\left( {0.1} \right)$
Using the change in base property (1) as in precious part, we get
${\log _{9\sqrt 3 }}\left( {0.1} \right) = \dfrac{{{{\log }_{10}}\left( {0.1} \right)}}{{{{\log }_{10}}\left( {9\sqrt 3 } \right)}} = \dfrac{{{{\log }_{10}}\left( {0.1} \right)}}{{{{\log }_{10}}\left( {{3^2}{{.3}^{\dfrac{1}{2}}}} \right)}} = \dfrac{{{{\log }_{10}}\left( {0.1} \right)}}{{{{\log }_{10}}\left( {{3^{\dfrac{5}{2}}}} \right)}}$
Using the property, ${\log _b}{x^c} = c{\log _b}x$in above, we get
\[{\log _{9\sqrt 3 }}\left( {0.1} \right) = \dfrac{{{{\log }_{10}}\left( {{{10}^{ - 1}}} \right)}}{{{{\log }_{10}}\left( {{3^{\dfrac{5}{2}}}} \right)}} = \dfrac{{ - 1.{{\log }_{10}}\left( {10} \right)}}{{\dfrac{5}{2}{{\log }_{10}}\left( 3 \right)}}\]
as stated above, ${\log _b}b = 1$, using this in above with $b = 10$, we get,
\[{\log _{9\sqrt 3 }}\left( {0.1} \right) = \dfrac{{ - 1}}{{\dfrac{5}{2}{{\log }_{10}}\left( 3 \right)}} = \dfrac{{ - 2}}{{5.{{\log }_{10}}\left( 3 \right)}}\]
Now we need to look for the value of \[{\log _{10}}\left( 3 \right)\]in the common log table which comes out to be (approx.) 0.477
\[ \Rightarrow {\log _{9\sqrt 3 }}\left( {0.1} \right) = \dfrac{{ - 2}}{{5 \times 0.477}} = - 0.838\]
Hence the value of ${\log _{9\sqrt 3 }}\left( {0.1} \right)$is (approx.) -0.838.

Note: The common logarithm is generally known as the logarithm with base 10. The natural logarithm of a number is its logarithm to the base of the mathematical constant e. Effort should be made to transform the number, of which log with base b is to be evaluated, in the form of base b. The value of logarithm of zero to any base is not defined. One may have to refer to the common lag table for value which could not be simplified with properties of log.