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

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.