Question

If \[{\log _2}x + {\log _4}x + {\log _{16}}x = \dfrac{{21}}{4}\] then find the value of x.
A) \[10\]
B) \[9\]
C) \[8\]
D) \[7\]

Answer 1

Hint: In this problem we have to solve the given equation by using the logarithm formula we have,
\[{\log _a}x = \dfrac{{{{\log }_c}x}}{{{{\log }_c}a}}\]
Simplifying the higher term till get the required term to solve the given equation for x. Then we get the value of x.

Complete step by step answer:
It is given in the question that, \[{\log _2}x + {\log _4}x + {\log _{16}}x = \dfrac{{21}}{4}\].
Now let us consider the first term and on applying logarithm formula to that term, we get,
\[{\log _2}x = \dfrac{{{{\log }_{10}}x}}{{{{\log }_{10}}2}}\]...... (1)
Let the above equation be equation (1)
Now let us consider the second term and on applying logarithm formula to that term, we get,
\[{\log _4}x = \dfrac{{{{\log }_{10}}x}}{{{{\log }_{10}}4}}\]...... (2)
Let the above equation be equation (2)
Now let us consider the third term and on applying logarithm formula to that term, we get,
\[{\log _{16}}x = \dfrac{{{{\log }_{10}}x}}{{{{\log }_{10}}16}}\]...... (3)
Let the above equation be equation (3)
Here let we add the equations (1), (2) and (3) we have,
\[{\log _2}x + {\log _4}x + {\log _{16}}x = \dfrac{{{{\log }_{10}}x}}{{{{\log }_{10}}2}} + \dfrac{{{{\log }_{10}}x}}{{{{\log }_{10}}4}} + \dfrac{{{{\log }_{10}}x}}{{{{\log }_{10}}16}}\]
Since 4 and 16 can be written as the powers of 2 the above equation is rewritten as follows,
\[{\log _2}x + {\log _4}x + {\log _{16}}x = \dfrac{{{{\log }_{10}}x}}{{{{\log }_{10}}2}} + \dfrac{{{{\log }_{10}}x}}{{{{\log }_{10}}{2^2}}} + \dfrac{{{{\log }_{10}}x}}{{{{\log }_{10}}{2^4}}}\]
Using one of the logarithmic identities we can write the above equation as,
\[{\log _2}x + {\log _4}x + {\log _{16}}x = \dfrac{{{{\log }_{10}}x}}{{{{\log }_{10}}2}} + \dfrac{{{{\log }_{10}}x}}{{2{{\log }_{10}}2}} + \dfrac{{{{\log }_{10}}x}}{{4{{\log }_{10}}2}}\]
Now on taking the terms that are in common we will arrive at the equation which follows,
\[{\log _2}x + {\log _4}x + {\log _{16}}x = \dfrac{{{{\log }_{10}}x}}{{{{\log }_{10}}2}}(1 + \dfrac{1}{2} + \dfrac{1}{4})\]
On solving the above equation we have,
\[{\log _2}x + {\log _4}x + {\log _{16}}x = \dfrac{{{{\log }_{10}}x}}{{{{\log }_{10}}2}}(\dfrac{{4 + 2 + 1}}{4}) = \dfrac{7}{4}\dfrac{{{{\log }_{10}}x}}{{{{\log }_{10}}2}}\]
Now let us substitute the value of\[{\log _2}x + {\log _4}x + {\log _{16}}x\] in the given equation, we get,
\[\dfrac{7}{4}\dfrac{{{{\log }_{10}}x}}{{{{\log }_{10}}2}} = \dfrac{{21}}{4}\]
On solving the above equation we arrive at the following step,
\[\dfrac{{{{\log }_{10}}x}}{{{{\log }_{10}}2}} = \dfrac{{21}}{4} \times \dfrac{4}{7} = 3\]
Now let us cross multiply the values in the equation, then we get,
\[{\log _{10}}x = 3{\log _{10}}2\]
Again using one of the logarithmic identity we get,
\[{\log _{10}}x = {\log _{10}}{2^3}\]
On substituting the value of cube of 2 we get,
\[{\log _{10}}x = {\log _{10}}8\]
By comparing both sides of the equation we get,
\[x = 8\]

Hence we have found that x=8 that is option (C) is the correct option.

Note:
We know that, \[{\log _a}{x^n} = n{\log _a}x\] which is the logarithmic identity used in the problem.
When two logarithmic functions with the same base are equal then we can equate the number.