Question

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

Answer 1

Hint: At first let's factorize the given number 1728 to write in the form of the base by using the properties ${a^n}*{b^n} = {\left( {ab} \right)^n}$ and by using the properties $\log {a^n} = n\log a$ and ${\log _a}a = 1$we get the required value.

Complete step-by-step answer:
We are given a number 1728
Now lets factorize the given number
$ \Rightarrow 1728 = 64*27 \\
   \Rightarrow 1728 = {2^6}*{3^3} \\
$
Now let's write the equation in a way that the powers of the numbers are the same
$ \Rightarrow 1728 = {2^6}*{\left( {\sqrt 3 } \right)^6}$
We know that ${a^n}*{b^n} = {\left( {ab} \right)^n}$
Therefore we have
$ \Rightarrow 1728 = {\left( {2\sqrt 3 } \right)^6}$
Now lets substitute this value in our given question
$ \Rightarrow {\log _{2\sqrt 3 }}1728 = {\log _{2\sqrt 3 }}{(2\sqrt 3 )^6}$
We know that $\log {a^n} = n\log a$
$ \Rightarrow {\log _{2\sqrt 3 }}1728 = 6{\log _{2\sqrt 3 }}(2\sqrt 3 )$
By using the property ${\log _a}a = 1$
We get,
$ \Rightarrow {\log _{2\sqrt 3 }}1728 = 6(1) = 6$
Therefore the value of ${\log _{2\sqrt 3 }}1728 = 6$.

Note: Points to remember while solving these types of problems:-
The term ${\log _b}n$ is well defined if
The base b is positive, and not equal to 1.
The number N is positive.
The log of 1 to any base is 0