Question

How do you find the inverse of \[f\left( x \right) = \dfrac{{100}}{{1 + {2^{ - x}}}}\] ?

Answer 1

Hint: This question involves the arithmetic operations like, addition/ subtraction/ multiplication/ division. We need to know the relation between exponential functions and logarithmic functions. Also, we need to know how to multiply a variable with variable, variable with constant, and constant with constant to make easy calculations. We need to know how to perform the arithmetic operations with different signs.

Complete step-by-step answer:
The given equation is shown below,
 \[f\left( x \right) = \dfrac{{100}}{{1 + {2^{ - x}}}}\]
We know that \[y\] is the function of \[x\] . So, the above equation can also be written as,
 \[y = \dfrac{{100}}{{1 + {2^{ - x}}}}\]
To find the inverse of \[y\] , we flip \[x\] and \[y\] . So, we get
 \[x = \dfrac{{100}}{{1 + {2^{ - y}}}}\]
Let’s move the term \[\left( {1 + {2^{ - y}}} \right)\] from RHS to LHS of the above equation we get,
 \[x\left( {1 + {2^{ - y}}} \right) = 100\]
Let’s move the term \[x\] from LHS to RHS, we get
 \[\left( {1 + {2^{ - y}}} \right) = \dfrac{{100}}{x}\]
Let’s move the term \[1\] from LHS to RHS of the above equation we get,
 \[{2^{ - y}} = \dfrac{{100}}{x} - 1 \to \left( 1 \right)\]
We know that,
 \[\dfrac{a}{b} - c = \dfrac{{a - bc}}{b}\]
By using this formula we can write,
 \[\dfrac{{100}}{x} - 1 = \dfrac{{100 - \left( {1 \times x} \right)}}{x} = \dfrac{{100 - x}}{x}\]
Let’s substitute the above equation in the equation \[\left( 1 \right)\] , we get
 \[\left( 1 \right) \to {2^{ - y}} = \dfrac{{100}}{x} - 1\]
 \[{2^{ - y}} = \dfrac{{100 - x}}{x}\] \[ \to \left( 2 \right)\]
We know that,
If,
 \[
  {e^x} = y, \\
  \log {e^x} = \log y \Rightarrow x = {\log _e}y \\
 \]
By using the above equations, the equation \[\left( 2 \right)\] becomes
 \[\left( 2 \right) \to {2^{ - y}} = \dfrac{{100 - x}}{x}\]
Take \[\log \] on both sides,
 \[\log {2^{ - y}} = \log \left( {\dfrac{{100 - x}}{x}} \right)\]
 \[ - y = {\log _2}\left( {\dfrac{{100 - x}}{x}} \right)\]
So, the final answer is,
 \[y = - {\log _2}\left( {\dfrac{{100 - x}}{x}} \right)\]
So, the correct answer is “ \[y = - {\log _2}\left( {\dfrac{{100 - x}}{x}} \right)\] ”.

Note: Note that to find the inverse form of the equation, we would flip \[x\] and \[y\] . Also, this question describes the operation of addition/ subtraction/ multiplication/ division. Note that logarithmic functions are the inverse form of exponential functions. So, the multiplication of \[\log \] and \[e\] is \[1\] . Note that \[\log {e^1} = 1\] and if we have the equation like \[\log {e^x} = \log y\] it will be converted as \[x = {\log _e}y\] . This relation will be used to solve these types of questions.