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If \[{\log _x}\left( {\dfrac{4}{9}} \right) = \dfrac{{ - 1}}{2}\], then the value of \[x\].

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Last updated date: 25th Jun 2024
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Answer
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Hint:
Here, we will find the value of the variable at the base. First, we will convert the logarithmic function into the exponential function. Then we will use the exponent rules to find the value of the variable. Logarithmic function is a function which is of the form \[y\] equals the log of \[x\] with the base \[b\].

Formula Used:
We will use the following formulas:
Negative Exponent Rule: \[{\left( x \right)^{ - n}} = {\left( {\dfrac{1}{x}} \right)^n}\]
Power Rule: \[{\left( {{a^m}} \right)^n} = \left( {{a^{mn}}} \right)\]

Complete step by step solution:
We are given with a logarithmic function \[{\log _x}\left( {\dfrac{4}{9}} \right) = \dfrac{{ - 1}}{2}\].
We will express the function as an exponential function.
We know that \[{\log _b}M = N\] can be expressed as \[M = {b^N}\].
Then the given expression \[{\log _x}\left( {\dfrac{4}{9}} \right) = \dfrac{{ - 1}}{2}\] can be expressed as the form
 \[ \Rightarrow \left( {\dfrac{4}{9}} \right) = {x^{\left( {\dfrac{{ - 1}}{2}} \right)}}\]
Since the power is negative, the expression on the right hand side gets reciprocal to change the power positive.
 By using the negative exponent rule \[{\left( x \right)^{ - n}} = {\left( {\dfrac{1}{x}} \right)^n}\] , we get
 \[ \Rightarrow \left( {\dfrac{4}{9}} \right) = {\left( {\dfrac{1}{x}} \right)^{\left( {\dfrac{1}{2}} \right)}}\]
By squaring on both the sides, we get
 \[ \Rightarrow {\left( {\dfrac{4}{9}} \right)^2} = {\left[ {{{\left( {\dfrac{1}{x}} \right)}^{\left( {\dfrac{1}{2}} \right)}}} \right]^2}\]
By using the Power rule \[{\left( {{a^m}} \right)^n} = \left( {{a^{mn}}} \right)\], we get
 \[ \Rightarrow {\left( {\dfrac{4}{9}} \right)^2} = {\left( {\dfrac{1}{x}} \right)^{\left( {\dfrac{1}{2}} \right) \times 2}}\]
Simplifying the expression, we get
 \[ \Rightarrow \left( {\dfrac{{16}}{{81}}} \right) = {\left( {\dfrac{1}{x}} \right)^1}\]
 \[ \Rightarrow \left( {\dfrac{1}{x}} \right) = \left( {\dfrac{{16}}{{81}}} \right)\]
Again by taking reciprocal on both sides, we get
 \[ \Rightarrow x = \dfrac{{81}}{{16}}\]

Therefore, the value of \[x\] is \[\dfrac{{81}}{{16}}\].

Note:
In order to solve the question we must convert the logarithmic function into an exponential function. The logarithmic function is of two types, the first type is that both the sides of the equation have logarithms, then the arguments are equal. i.e.,
 \[\begin{array}{l}{\log _b}M = {\log _b}N\\ \Rightarrow M = N\end{array}\] .
The second type is that only one side of the equation has a logarithmic function, then the equation on the right becomes the exponent of the base of the logarithm. i.e.,
 \[\begin{array}{l}{\log _b}M = N\\ \Rightarrow M = {b^N}\end{array}\].
We should be careful in using the correct exponent rule at suitable places.