Question

How do you solve ${{\log }_{x}}\left( \dfrac{1}{8} \right)=-\dfrac{3}{2}$?

Answer 1

Hint: To solve the given expression we will use the properties of exponentiation and logarithm. First we will convert the given expression in exponentiation form by using the relation that ${{\log }_{a}}x=y$ is equivalent to ${{a}^{y}}=x$. By using this property we form an equation and simplify it to get the desired answer.

Complete step-by-step solution:

We have been given that ${{\log }_{x}}\left( \dfrac{1}{8} \right)=-\dfrac{3}{2}$.
We have to solve the given expression and find the value of x.
We know that logarithm is the inverse function to the exponentiation. The exponent of a number says how many times to use a number in multiplication. We will use the basic properties of logarithm to solve further.
We know that ${{\log }_{a}}x=y$ is equal to ${{a}^{y}}=x$.
So by applying the above property we will get
$\begin{align}
  & \Rightarrow {{\log }_{x}}\left( \dfrac{1}{8} \right)=-\dfrac{3}{2} \\
 & \Rightarrow {{x}^{\dfrac{-3}{2}}}=\dfrac{1}{8} \\
\end{align}$
Now, we know that ${{x}^{-a}}=\dfrac{1}{{{x}^{a}}}$
Applying the property to the above obtained equation we will get
\[\Rightarrow \dfrac{1}{{{x}^{\dfrac{3}{2}}}}=\dfrac{1}{8}\]
Now, simplifying the above equation we will get
$\Rightarrow {{x}^{\dfrac{3}{2}}}=8$
Now, we know that $8={{2}^{3}}$
Substituting the value we will get
$\Rightarrow {{x}^{\dfrac{3}{2}}}={{2}^{3}}$
Now, taking the square on both sides we will get
\[\Rightarrow {{\left( {{x}^{\dfrac{3}{2}}} \right)}^{2}}={{\left( {{2}^{3}} \right)}^{2}}\]
Now, we know that ${{\left( {{x}^{m}} \right)}^{n}}={{x}^{mn}}$
Applying the property to the above obtained equation we will get
\[\begin{align}
  & \Rightarrow \left( {{x}^{\dfrac{3}{2}\times 2}} \right)=\left( {{2}^{3\times 2}} \right) \\
 & \Rightarrow {{x}^{3}}=\left( {{2}^{3\times 2}} \right) \\
\end{align}\]
Now, taking the cube root both the sides we will get
$\begin{align}
  & \Rightarrow x={{2}^{2}} \\
 & \Rightarrow x=4 \\
\end{align}$
So on solving the given expression we get the value $x=4$.

Note: To solve such types of questions students must have the knowledge of properties of logarithm and properties of exponentiation. The possibility of mistake is while applying the property ${{\log }_{a}}x=y$. Students must carefully compare the values of a, x and y and substitute properly.