Question

For \[x > 0\] and \[x\] not equal to 1, \[{\log _{16}}\left( x \right) = \]

Answer 1

Hint:In order to solve the problem related to logarithm first we will use the concept and theorem of logarithm to simplify the given term. We will convert the base of the given problem as that of the solution and use the conditions to proceed further.

Complete step-by-step answer:
Given that:
\[x > 0\]
Also we have the condition
\[x\] not equal to 1
To solve:
\[{\log _{16}}\left( x \right)\]
As we can notice that the base of the logarithm of the given problem statement is 16 but the base of the logarithm of solution options is 2.
So we will simplify the problem term to bring the base as 2.
As we know that the formula relating the base of the logarithm is given as:
\[{\log _q}\left( p \right) = \dfrac{{{{\log }_e}\left( p \right)}}{{{{\log }_e}\left( q \right)}}\]
Let us now use the above formula in the problem logarithm term to change the base.
\[{\log _{16}}\left( x \right) = \dfrac{{{{\log }_e}\left( x \right)}}{{{{\log }_e}\left( {16} \right)}}\]
Now in order to get the final answer we shall modify the numerator and denominator and then further use the reverse of above formula.
First let us simplify the denominator.
As we know the formula relating the coefficient of logarithm and the term under logarithm is given by:
\[\log {p^q} = q\log p\]
Using the above formula in the denominator of the given term we get:
\[
   = \dfrac{{{{\log }_e}\left( x \right)}}{{{{\log }_e}\left( {16} \right)}} \\
   = \dfrac{{{{\log }_e}\left( x \right)}}{{{{\log }_e}\left( {{2^4}} \right)}} \\
   = \dfrac{{{{\log }_e}\left( x \right)}}{{4{{\log }_e}\left( 2 \right)}} \\
 \]
As we have the numerator and the denominator in the simplest form let us use the reverse of the formula used above to get the required term. So we have:
\[
   = \dfrac{{{{\log }_e}\left( x \right)}}{{4{{\log }_e}\left( 2 \right)}} \\
   = \dfrac{1}{4} \times \dfrac{{{{\log }_e}\left( x \right)}}{{{{\log }_e}\left( 2 \right)}} \\
   = \dfrac{1}{4} \times {\log _2}\left( x \right){\text{ }}\left[ {\because \log {p^q} = q\log p} \right] \\
   = 0.25 \times {\log _2}\left( x \right) \\
 \]
Hence, the given term is \[ = 0.25 \times {\log _2}\left( x \right)\]

Note:In order to solve such types of problems involving logarithm, students must remember all the formulas related to the logarithm function and the theorems. Some of these are mentioned above along with the solution. Students must not consider some specific cases in order to solve these types of problems and must proceed with general terms only.