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

How do you solve log⁡x(18)=−32{{\log }_{x}}\left( \dfrac{1}{8} \right)=-\dfrac{3}{2}?

Explanation

Solution

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⁡ax=y{{\log }_{a}}x=y is equivalent to ay=x{{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(18)=−32{{\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⁡ax=y{{\log }_{a}}x=y is equal to ay=x{{a}^{y}}=x.
So by applying the above property we will get
⇒log⁡x(18)=−32 ⇒x−32=18 \begin{aligned} & \Rightarrow {{\log }_{x}}\left( \dfrac{1}{8} \right)=-\dfrac{3}{2} \\\ & \Rightarrow {{x}^{\dfrac{-3}{2}}}=\dfrac{1}{8} \\\ \end{aligned}
Now, we know that x−a=1xa{{x}^{-a}}=\dfrac{1}{{{x}^{a}}}
Applying the property to the above obtained equation we will get
⇒1x32=18\Rightarrow \dfrac{1}{{{x}^{\dfrac{3}{2}}}}=\dfrac{1}{8}
Now, simplifying the above equation we will get
⇒x32=8\Rightarrow {{x}^{\dfrac{3}{2}}}=8
Now, we know that 8=238={{2}^{3}}
Substituting the value we will get
⇒x32=23\Rightarrow {{x}^{\dfrac{3}{2}}}={{2}^{3}}
Now, taking the square on both sides we will get
⇒(x32)2=(23)2\Rightarrow {{\left( {{x}^{\dfrac{3}{2}}} \right)}^{2}}={{\left( {{2}^{3}} \right)}^{2}}
Now, we know that (xm)n=xmn{{\left( {{x}^{m}} \right)}^{n}}={{x}^{mn}}
Applying the property to the above obtained equation we will get

& \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{aligned}$$ Now, taking the cube root both the sides we will get $\begin{aligned} & \Rightarrow x={{2}^{2}} \\\ & \Rightarrow x=4 \\\ \end{aligned}$ **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.