The summation of logarithm of $x$ to base $2$, logarithm of $x$ to base $4$ and logarithm of $x$ to base $16$ is equal to the quotient of $21$ and $4$.

$\log_{2} x + \log_{4} x + \log_{16} x = \dfrac{21}{4}$

The bases of three logarithmic terms are $2$, $4$ and $16$ and it is not possible to solve this equation until the bases of all three terms are converted as the every term has a common base. It is possible in this logarithm problem by expressing the bases of second and third logarithmic functions in exponential form in the form the base of first logarithmic term.

$(1) \,\,\,\,\,\,$ $\log_{2} x = \log_{2} x$

$(2) \,\,\,\,\,\,$ $\log_{4} x = \log_{2^2} x$

$(3) \,\,\,\,\,\,$ $\log_{16} x = \log_{2^4} x$

Substitute them in the given logarithmic equation in their associated terms.

$\implies \log_{2} x + \log_{2^2} x + \log_{2^4} x = \dfrac{21}{4}$

Use the power rule of logarithm to separate the exponent of the base from the logarithmic function.

$\implies \log_{2} x + \dfrac{1}{2} \log_{2} x + \dfrac{1}{4} \log_{2} x = \dfrac{21}{4}$

The logarithm of $x$ to base $2$ is the common term in the left hand side of this equation. So, take $\log_{2} x$ common from them.

$\implies \Bigg[1 + \dfrac{1}{2} + \dfrac{1}{4}\Bigg] \log_{2} x = \dfrac{21}{4}$

$\implies \Bigg[\dfrac{4+2+1}{4}\Bigg] \log_{2} x = \dfrac{21}{4}$

$\implies \dfrac{7}{4} \times \log_{2} x = \dfrac{21}{4}$

$\implies \log_{2} x = \dfrac{21}{4} \times \dfrac{4}{7} $

$\implies \log_{2} x = \dfrac{21 \times 4}{4 \times 7}$

$\implies \require{cancel} \log_{2} x = \dfrac{\cancel{21} \times \cancel{4}}{\cancel{4} \times \cancel{7}}$

$\implies \log_{2} x = 3$

$\implies x = 2^3$

$\therefore \,\,\,\,\,\, x = 8$

Therefore, it is derived that the value of $x$ is $8$ and it is the required solution for this logarithm problem.

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