Question

Find the value of a, if ${\log _a}\sqrt x = 4$
A) ${x^4}$
B) ${x^{\dfrac{1}{4}}}$
C) ${x^2}$
D) ${x^{\dfrac{1}{8}}}$

Answer 1

In this question, we have to find the value of $a$. So, the concept is to apply the basic logarithmic. The given expression, ${\log _a}\sqrt x = 4$ is in the form of identity ${\log _b}y = x$ and it can be written as ${b^x} = y$ form. And we also use the exponent law of power ${\left( {{y^m}} \right)^n} = {y^{m \times n}}$ to find the answer.

Complete step by step answer:
For finding the value of $a$, simplifying the given equation ${\log _a}\sqrt x = 4$.
The given expression is in the form of ${\log _b}y = k$ and it can be written as ${b^k} = y$
So, comparing the given equation from the identity we find that
$\Rightarrow b = a$, $y = \sqrt x$ and $k = 4$
Therefore, the given equation can be written as
$\Rightarrow {a^4} = \sqrt x$
We know that an equation can be raised to the same power on both sides without altering its value. Thus, raising the power of $\dfrac{1}{4}$ on both sides of the above equation, we’ll get:
$\Rightarrow {a^{\dfrac{4}{4}}} = {\left( {\sqrt x } \right)^{\dfrac{1}{4}}}$
Square root means the power of $\dfrac{1}{2}$, putting this in the above equation, we’ll get
$\Rightarrow a = {\left( {{x^{\dfrac{1}{2}}}} \right)^{\dfrac{1}{4}}}$
Further, from the exponent law of power we know that ${\left( {{y^m}} \right)^n} = {y^{m \times n}}$. Therefore we have:
$\Rightarrow a = {x^{\dfrac{1}{2} \times \dfrac{1}{4}}} \\\ \Rightarrow a = {x^{\dfrac{1}{8}}} \\\$

Hence, option $\left( D \right)$ is correct.

Additional information:
There are mainly two types of logarithm which we study, one is the logarithm of the base $10$ that is a common logarithm and the second is the logarithm of base $e$ that is a natural logarithm. We also study the logarithm of the base of any other whole number than $10$ and $e$. The logarithm of any negative number does not exist.

Note:
Some other properties of logarithm are:
\Rightarrow \log m + \log n = \log mn \\\ \Rightarrow \log m - \log n = \log \dfrac{m}{n} \\\ \Rightarrow a\log m = \log {m^a} \\\
Logarithm problems are solved by frequently using these properties.