Question

How do you solve $x={{\log }_{36}}6$ ?

Answer 1

As we can see that the above equation is a logarithmic equation. In order to solve it we have to apply basic rules of logarithms. First, we will try to condense the log expression into simple logarithms then we will use the rules to isolate the logarithmic expressions which have the same bases on both sides of the equation. Our next step is to set the arguments equal to each other and then simplify or solve the resulting equation.

Complete answer:
This question belongs to the concept of solving logarithmic equations or functions. Logarithmic equations involve logarithm of an expression logarithm is just the opposite or inverse of exponentiation. Thus, we can conclude that the logarithm of a given function is the exponent to which another number must be raised in order to get the original number.
Now in the question we have $x={{\log }_{36}}6$ . Here we can see that the expression is a simple logarithmic equation therefore in order to solve the given logarithmic equation we will go with the basic definition of logarithms.
First, we will rewrite the given equation in exponential form as per the definition of logarithm states that if x and z are positive real numbers also z is not equal to zero then $b={{\log }_{z}}x$ can be written as ${{z}^{b}}=x$
Therefore,

& x={{\log }_{36}}6 \\\ & \Rightarrow {{36}^{x}}=6 \\\ \end{aligned}$$ Now applying basic exponential rules to simplify the above equation. $$\begin{aligned} & \Rightarrow {{({{6}^{2}})}^{x}}=6 \\\ & \Rightarrow {{6}^{2}}^{x}=6 \\\ \end{aligned}$$ Now comparing powers on both sides, we get, $$\begin{aligned} & 2x=1 \\\ & \Rightarrow x=\dfrac{1}{2} \\\ \end{aligned}$$ Therefore, the value of x is $$\dfrac{1}{2}$$ . **Note:** We can solve this question using antilog but for that we have to use logarithmic tables. While solving the question keep in mind that logarithmic and exponential rules, both are different. Also, while calculating the value of x, equate the correct values. We will never get a negative value after solving a logarithmic equation.