Question

The inverse of \(\left[\left(7^{x}-7^{-x}\right) /\left(7^{x}+7^{-x}\right)\right]\) is (a) \((1 / 2) \log _{7}[(1+\mathrm{x}) /(1-\mathrm{x})]\) (b) \(\log _{7}[(1-x) /(1+x)]\) (c) \(\log _{(1 / 2)}[(1-x) /(1+x)]\) (d) \((1 / 2) \log _{e}[(1+x) /(1-x)]\)

Short answer

The derived inverse function is \(f^{-1}(x) = \log_7{\frac{x7^{2y}}{x+1}}\). However, none of the given options match this result.

Step-by-step solution

Rewrite the function using properties of exponents

Express the function \(f(x)\) as follows: \(f(x) = \frac{7^{x}-7^{-x}}{7^{x}+7^{-x}}=\frac{7^x\frac{1-7^{-2x}}{1+7^{-2x}}}\) Now, let \(y = f(x)\).

Find the inverse of the function

To find the inverse of the function, first express \(x\) in terms of \(y\) and then replace \(y\) with \(f^{-1}(x)\) : \(y=\frac{7^x\frac{1-7^{-2x}}{1+7^{-2x}}}\) Let's isolate \(7^x\): \(y(1+7^{-2x})=7^x(1-7^{-2x})\) \(y+ y (7^{-2x})=7^{x}(1-7^{-2x})\) Now multiply both sides by \(7^{2x}\): \(y (7^{2x})+y(1)=7^{x}(7^{2x})(1-7^{-2x})\) \(y(7^{2x})+y=7^{3x}-7^x\) Now, we group terms and get the following: \((y+1)7^x = y7^{2x}\) Now we obtain \(7^x\): \(7^x = \frac{y7^{2x}}{y+1}\) Now, we can take the logarithm to base 7: \(x = \log_7{\frac{y7^{2x}}{y+1}}\) Finally, replace \(y\) with \(f^{-1}(x)\) and \(x\) with \(y\): \(f^{-1}(x) = \log_7{\frac{x7^{2y}}{x+1}}\)

Compare the derived inverse function with the given options

The derived inverse function is \(f^{-1}(x) = \log_7{\frac{x7^{2y}}{x+1}}\). Unfortunately, none of the given options match the derived inverse function.

Key concepts

Properties of Exponents

Understanding the properties of exponents is crucial for solving exponential equations and finding inverses. Exponents are mathematical symbols that represent repeated multiplication of a base number. For example, \( a^n \) means you multiply the base \( a \) by itself \( n \) times.
When working with exponents, there are several important rules or properties to remember:

• Product of Powers: \( a^m \times a^n = a^{m+n} \). This property states that when you multiply two exponents with the same base, you add the exponents.


• Quotient of Powers: \( \frac{a^m}{a^n} = a^{m-n} \). When you divide two exponents with the same base, you subtract the exponents.


• Power of a Power: \( (a^m)^n = a^{mn} \). This implies that raising an exponent to another power multiplies the exponents.


• Zero Exponent: \( a^0 = 1 \), assuming the base \( a \) is not zero. A number raised to the power of zero equals one.


• Negative Exponent: \( a^{-n} = \frac{1}{a^n} \). A negative exponent indicates reciprocal.

Use these properties to simplify and manipulate expressions as shown in the solution to find the inverse function. Recognizing and using these properties can help to rewrite functions, making it easier to solve for variables involved.

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions. This means if you have an exponential equation, turning it into a logarithmic equation can help solve for the unknown. Logarithms are useful for finding inverses because they allow you to "undo" the exponent.

For logarithms:

• Definition: A logarithm \( \log_b(a) \) answers the question: "To what exponent do we raise \( b \) to get \( a \)?" So if \( b^c = a \), then \( \log_b(a) = c \).


• Common Logarithms: If the base is 10, it’s often written as \( \log(a) \). This is called a common logarithm.


• Natural Logarithms: If the base is \( e \) (where \( e \approx 2.718 \)), it is written as \( \ln(a) \). This is called a natural logarithm.


• Logarithmic Properties:
  • \( \log_b(mn) = \log_b(m) + \log_b(n) \)
  • \( \log_b(\frac{m}{n}) = \log_b(m) - \log_b(n) \)
  • \( \log_b(m^n) = n \log_b(m) \)
  • \( \log_b(b) = 1 \)
  • \( \log_b(1) = 0 \)

When dealing with inverse functions of an exponential form, logarithmic functions are used to express terms in a manageable way, just like in the solution where they converted expressions using \( \log_7 \). This simplifies solving for the desired variable.

Exponential Equations

Exponential equations are types of equations where variables appear as exponents. Solving exponential equations often involves rewriting these equations using simpler terms, like logarithms or properties of exponents.

To solve exponential equations:

• Equating Powers: If you have an equation like \( a^x = a^y \), it's simple: \( x = y \) as long as the bases are the same and not zero.


• Using Logarithms: Sometimes, equating powers directly isn't possible. You use logarithms to help solve the equation. For example, \( b^x = a \), apply a logarithm to get \( x = \log_b(a) \).


• Isolating the Exponential Function: It’s often necessary to isolate the exponential part of the equation before applying logarithms. Simplifying sides of equations (e.g., factor out common terms) can make your life easier.

In our original exercise, the approach involved transforming the function through applying logarithms to directly solve and derive the inverse function. Understanding and applying these methods helps to achieve solutions to even the intricate inverse problems efficiently.