Question

Find the value of $\log _{2\sqrt 3 }^{1728}$.

Answer 1

Hint: This is a problem of logarithm and we have to find the value of $\log _{2\sqrt 3 }^{1728}$. Here base is $2\sqrt 3 $. Its value can be calculated by using a formula of logarithm that is $\log _a^{{a^n}} = n$. Firstly, we have to write the prime factor of $1728$ then arrange the factors in the form of ${\left( {2\sqrt 3 } \right)^x}$ where $x$ is a rational number. then apply the above formula to get the required result.

Complete step-by-step answer:
Given: to find the value of $\log _{2\sqrt 3 }^{1728}$.
Here, the base of logarithm is $2\sqrt 3 $ .
Now, we have to write the prime factors of $1728$.
\[
  \left. 2 \right|\underline {1728} \\
  \left. 2 \right|\underline {864} \\
  \left. 2 \right|\underline {432} \\
  \left. 2 \right|\underline {216} \\
  \left. 2 \right|\underline {108} \\
  \left. 2 \right|\underline {54} \\
  \left. 3 \right|\underline {27} \\
  \left. 3 \right|\underline 9 \\
  \,\,\,3 \\
 \]
The prime factor of $1728$ is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$.
The prime factor of $1728$ can be written as ${2^6} \times {3^3} = {2^6} \times {3^{\dfrac{6}{2}}} = {2^6} \times {\left( {\sqrt 3 } \right)^6} = {\left( {2\sqrt 3 } \right)^6}$.
Now, $\log _{2\sqrt 3 }^{1728}$ can be written as $\log _{2\sqrt 3 }^{{{\left( {2\sqrt 3 } \right)}^6}}$.
By applying the above given formula $\log _a^{{a^n}} = n$. we get the value of $\log _{2\sqrt 3 }^{1728}$ is $6$.

Thus, the required value of $\log _{2\sqrt 3 }^{1728}$ is $6$.

Note:
Some important formulas of logarithm which may be used solved various problems of logarithm.
(1) $\log _{{a^n}}^{{b^m}} = \dfrac{m}{n}\log _a^b$
The value of the above given question can also be calculated by using this formula. The base $2\sqrt 3 $ can be written as ${\left( {{{\left( {2\sqrt 3 } \right)}^2}} \right)^{\dfrac{1}{2}}} = {\left( {12} \right)^{\dfrac{1}{2}}}$ and the after finding the prime factors we can write $1728$ as ${\left( {12} \right)^3}$. Then, applying this formula we can write $\log _{2\sqrt 3 }^{1728}$ as $\dfrac{3}{{\dfrac{1}{2}}}\log _{12}^{12} = \dfrac{{3 \times 2}}{1}\log _{12}^{12} = 6$.
(2) $\log _a^{b \times c} = \log _a^b + \log _a^c$.
(3) $\log _a^{\dfrac{b}{c}} = \log _a^b - \log _a^c$.
(4) $\log _a^b \times \log _b^c \times \log _c^d = \log _a^d$.
The fourth formula is beneficial in solving the problem when the base of logarithm is equal to the function.
There is a basic difference between $\log x$ and $\ln x$. $\ln x$ is called a natural logarithm and its base is equal to $e$, whereas $\log x$ is a logarithm whose base is generally $10$.