For the exponential function f, the value of f(1) is k, where k is a constant. Which of the following equivalent forms of the function f displays k as the coefficient or the base? 1. f(x) = 50(1.6)^(x+1) 2. f(x) = 80(1.6)^(x) 3. f(x) = 128(1.6)^(x-1) 4. f(x) = 204.8(1.6)^(x-2)

Mathematics Questions

For the exponential function f, the value of f(1) is k, where k is a constant. Which of the following equivalent forms of the function f displays k as the coefficient or the base? 1. f(x) = 50(1.6)^(x+1) 2. f(x) = 80(1.6)^(x) 3. f(x) = 128(1.6)^(x-1) 4. f(x) = 204.8(1.6)^(x-2)

Short Answer

The problem involves identifying an exponential function that reveals the constant k when x equals 1. After evaluating the provided functions, it is determined that f(x) = 128(1.6)^(x-1) satisfies this requirement, clearly displaying k as 128.

Step-by-Step Solution

Step 1: Understand the Problem

The problem requires identifying which exponential function allows the constant k to be displayed as either the coefficient or the base when the input x equals 1. We are specifically looking for a function that simplifies to k when evaluated at x = 1.

Step 2: Evaluate Each Function

We will substitute x = 1 into each of the given functions to check which function directly shows k. The functions to evaluate are:

f(x) = 50(1.6)^(x+1)
f(x) = 80(1.6)^(x)
f(x) = 128(1.6)^(x-1)
f(x) = 204.8(1.6)^(x-2)

By evaluating these, we find that only one of them yields k as a clear output.

Step 3: Identify the Correct Function

After evaluating, we find:

For f(x) = 128(1.6)^(x-1), substituting x=1 gives f(1) = 128(1.6)^0 = 128, directly showing k.
The other functions either change k through multiplication with the base or exponent, failing to display it directly.

Therefore, the function that displays k clearly is f(x) = 128(1.6)^(x-1).

For the exponential function f, the value of f(1) is …