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

In the given logarithmic equation, the left hand side expression contains three logarithmic terms, which are logarithm of $x$ to base $2$, the logarithm of $x$ to base $4$ and the logarithm of $x$ to base $16$ but the sum of them is equal to a rational number $\large \frac{21}{4}$, which is expressed in right side expression.

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

Convert the log terms to required form

The bases of three logarithmic terms are $2$, $4$ and $16$. Due to the different bases, it is not possible to solve this equation until we make them same. Fortunately, it is possible to express the bases of the second and third logarithmic functions in the form of the base of first logarithmic term by exponentiation in this logarithm problem.

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

Now, substitute them in the given logarithmic equation at their respective places.

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

Simplify the logarithmic equation

According to the base power rule of logarithm, the exponents can be separated from the bases of the logarithmic functions at base position.

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

Solve the Logarithmic equation

The logarithm of $x$ to base $2$ is a common factor in every term of the left hand side of the equation. So, it can be taken out common from them.

$\implies$ $\Bigg(1+\dfrac{1}{2}+\dfrac{1}{4}\Bigg)\log_{2}{x}$ $\,=\,$ $\dfrac{21}{4}$

The numerical factor is a sum of three rational numbers in the left hand side of the equation. It can be simplified by using the addition of rational numbers.

$\implies$ $\Bigg(\dfrac{4+2+1}{4}\Bigg)\log_{2}{x}$ $\,=\,$ $\dfrac{21}{4}$

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

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

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

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

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

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

Now, use the relationship between logarithms and exponents for expressing the logarithmic equation as an exponential equation.

$\implies$ $x \,=\, 2^3$

$\implies$ $x \,=\, 2 \times 2 \times 2$

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

Therefore, it is solved that the value of $x$ is equal to $8$.