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.

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

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.

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

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

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

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 $8$.

Latest Math Topics

Latest Math Problems

Email subscription

Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers.
Know more

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.