Question

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The inverse function of the exponential function f(x) = a^x is the (a. transcendental, b. logarithmic, c. rational, d. polynomial, e. algebraic) function with base a.

Answer 1

The inverse function of the exponential function f(x) = a^x is the logarithmic function with base a.

The inverse function of a function f is a function that "undoes" the original function, meaning that it reverses the transformation applied by the original function. In the case of the exponential function, the inverse function is the logarithmic function, which "undoes" the transformation applied by the exponential function.

For example, suppose we have the exponential function f(x) = 2^x. The inverse function of this function is the logarithmic function with base 2, which is written as y = log_2 x. If we apply the inverse function to 2^x, we get:

y = log_2 (2^x)

Solving for x gives:

x = 2^y

This means that the inverse function of the exponential function f(x) = a^x is the logarithmic function with base a, which is written as y = log_a x.

The other options (a. transcendental, c. rational, d. polynomial, e. algebraic) are not correct, since they do not describe the inverse function of the exponential function.

Answer 2

b. logarithmic

Explanation:

Given exponential function:

`f(x)=a^x`

The inverse of the given function is the logarithmic function with base a.

To find the inverse of a function, replace f(x) with y:

`\implies y=a^x`

Swap the x and y:

`\implies x=a^y`

Take logs with base a of both sides of the equation:

`\implies \log_ax=\log_aa^y`

Apply the log power law: logₐ xⁿ = n logₐ x

`\implies \log_ax=y\log_aa`

Apply the log law: logₐ a = 1

`\implies \log_ax=y`

`\implies y=\log_ax`

Replace y with f⁻¹(x):

`\implies f^(-1)(x)=\log_ax`

Thus proving that the inverse of the exponential function f(x) = aˣ is the logarithmic function with base a.