Question

How do you solve ${\log _2}\left( {\dfrac{x}{2}} \right) = 5$?

Answer 1

Hint: In this question, we need to simplify the given logarithmic function and obtain the solution. We know that the logarithm functions are the inverse of exponential functions. In exponential function, one term is raised to the power of another term, i.e. of the form $x = {b^y}$ is an exponential function and the inverse of this function is $y = {\log _b}x$ which is a logarithmic function. Since in this question we are given a logarithmic function, we convert it into exponential function using the rule explained. And then we simplify it to obtain the value of the variable x.

Complete step by step answer:
Given the logarithmic function ${\log _2}\left( {\dfrac{x}{2}} \right) = 5$ …… (1)
We are asked to solve the logarithmic function in the equation (1) to obtain the solution.
We know that logarithm is the inverse of exponential function. Note that logarithm form and index (exponential) form are interchangeable.
i.e. if ${b^y} = x$, then we have ${\log _b}x = y$ …… (2)
where log denotes the logarithmic function.
Here x is an argument of logarithm function which is always positive.
And b is called the base of logarithm function.
Using the concept mentioned above, we can solve the given equation and express it in exponential form and then simplify it.
Now consider the given function given in the equation (1).
Now comparing it with the general form given in the equation (2).
We have here $b = 2$ and $y = 5$
Thus comparing we get,
${\log _2}\left( {\dfrac{x}{2}} \right) = 5$
$ \Rightarrow {2^5} = \dfrac{x}{2}$
Multiplying both sides by 2, we get,
$ \Rightarrow 2 \times {2^5} = \dfrac{x}{2} \times 2$
This can also be written as,
$ \Rightarrow {2^1} \times {2^5} = \dfrac{x}{2} \times 2$ …… (3)
We know that ${x^a} \cdot {x^b} = {x^{a + b}}$
Applying this to the L.H.S. of the equation (3).
We have here $x = 2$, $a = 1$ and $b = 5$
Hence we get,
$ \Rightarrow {2^1} \cdot {2^5} = {2^{1 + 5}}$
$ \Rightarrow {2^1} \cdot {2^5} = {2^6}$
Now simplifying the above expression in the equation (2), we get,
$ \Rightarrow {2^6} = x$
$ \Rightarrow 64 = x$
Hence we get $x = 64$.

Thus, the solution for the expression ${\log _2}\left( {\dfrac{x}{2}} \right) = 5$ is given by $x = 64$.

Note: If the question has the word log or $\ln $, it represents the given function as logarithmic function. Note that we have two types of logarithmic function.
One is a common logarithmic function which is represented as a log and its base is 10.
The other one is a natural logarithmic function represented as $\ln $and its base is $e$.
We must know the basic properties of logarithmic functions and note that these properties hold for both log and $\ln $functions.
Some properties of logarithmic functions are given below.
(1) $\log (x \cdot y) = \log x + \log y$
(2) $\log \left( {\dfrac{x}{y}} \right) = \log x - \log y$
(3) $\log {x^n} = n\log x$
(4) $\log 1 = 0$
(5) ${\log _e}e = 1$