Question

The value of \[{{\log }_{2\sqrt{3}}}1728\] is equal to
a) 3
b) 4
c) 5
d) 6

Answer 1

Hint: First convert the term \[{{\log }_{c}}d\] to the fraction of \[\dfrac{\log d}{\log c}\], then change the number d, c in form of \[{{a}^{b}}\], where a is the small possible prime factor of both c, d then make b according to it. Then do further calculation and get the results.

Complete step-by-step answer:

In the question we are asked to find the value of \[{{\log }_{2\sqrt{3}}}1728\].

For evaluating this we have to know the fact that \[{{\log }_{a}}b\] can be written or represented as, \[\dfrac{\log b}{\log a}\].

So, we can represent \[{{\log }_{2\sqrt{3}}}1728\] as,

\[\dfrac{\log 1728}{\log 2\sqrt{3}}\] or \[\dfrac{\log 1728}{\log \sqrt{12}}\]

Now, use the formula or the transformation rule which is \[{{\log }_{a}}b\] which can be further written as \[b\log a\], where b can be any real number whether rational or irrational number.

Now, we will represent each of the numerator and denominator of the fraction, \[\dfrac{\log 1728}{\log \sqrt{12}}\] as in form of \[{{\log }_{a}}b\] such that their powers may vary but base remains same.

So, we will represent both 1728 and \[\sqrt{12}\] in powers of 12.

So, we can represent the number 1728 as in powers of 12. So, the number 1728 can be written as \[{{12}^{3}}\].

So, \[\log 1728\] can also be written as \[\log {{12}^{3}}\]. Hence, using \[\log {{a}^{b}}\]

transformation, while transforming into \[b\log a\] we get, \[\log {{12}^{3}}\] as \[3\log 12\].

Hence, \[\log 1728\] can be written as \[3\log 12\].

So, now we will represent the \[\sqrt{12}\] as in powers of 12. So, the \[\sqrt{12}\] can be written as \[{{12}^{\dfrac{1}{2}}}\].

So, \[\log \sqrt{12}\] can also be written as \[\log {{12}^{t}}\]. Hence, using \[\log {{a}^{b}}\] trans which transforms into \[b\log a\] we get,

\[\log {{12}^{\dfrac{1}{2}}}\] as \[\dfrac{1}{2}\log 12\]

Hence, \[\log \sqrt{12}\] can be written as \[\dfrac{1}{2}\log 12\].

Now, we will write the fraction,

\[\dfrac{\log 1728}{\log \sqrt{12}}\] or \[\dfrac{3\log 12}{\dfrac{1}{2}\log 12}\]

Now, by cancelling \[\log 12\] from both numerator and denominator and doing the calculation we get,

\[\dfrac{\log 1728}{\log \sqrt{12}}=\dfrac{3}{\dfrac{1}{2}}=6\]

Hence, the answer is (d).

Note: Students while doing these kinds of problems while changing numbers into form \[{{a}^{b}}\] they should take a as the smallest possible prime factor to avoid any problems in the solution in further steps and should also be careful about the calculation mistakes.