Select the correct answer.
Using a table of values, determine the solution to the equation below to the nearest fourth of a unit.
[tex]$2^{(-x)}+1=5^x+2$[/tex]
A. [tex]$x=2.50$[/tex]
B. [tex]$x=0.75$[/tex]
C. [tex]$x=-0.50$[/tex]
D. [tex]$x=-1.50$[/tex]
Answer (1)
Rewrite the equation as 2 − x − 5 x = 1 .
Evaluate f ( x ) = 2 − x − 5 x for the given values of x: 2.50, 0.75, -0.50, -1.50.
Check which value of x makes f ( x ) closest to 1.
Since f ( − 0.50 ) ≈ 0.9670 is the closest to 1, the solution is approximately x = − 0.50 .
Explanation
Understanding the Problem We are given the equation 2 − x + 1 = 5 x + 2 and asked to find the solution to the nearest fourth of a unit using a table of values. The given options are A. x = 2.50 , B. x = 0.75 , C. x = − 0.50 , D. x = − 1.50 .
Rewriting the Equation and Defining a Function First, let's rewrite the equation as 2 − x − 5 x = 1 . We want to find the value of x from the given options that satisfies this equation. We can define a function f ( x ) = 2 − x − 5 x and evaluate it for each of the given x values. We are looking for the value of x for which f ( x ) is closest to 1.
Evaluating the Function Now, let's evaluate f ( x ) for each of the given options:
For A, x = 2.50 : f ( 2.50 ) = 2 − 2.50 − 5 2.50 ≈ 0.1768 − 55.9017 ≈ − 55.7249 For B, x = 0.75 : f ( 0.75 ) = 2 − 0.75 − 5 0.75 ≈ 0.5946 − 3.3437 ≈ − 2.7491 For C, x = − 0.50 : f ( − 0.50 ) = 2 − ( − 0.50 ) − 5 − 0.50 = 2 0.50 − 5 − 0.50 ≈ 1.4142 − 0.4472 ≈ 0.9670 For D, x = − 1.50 : f ( − 1.50 ) = 2 − ( − 1.50 ) − 5 − 1.50 = 2 1.50 − 5 − 1.50 ≈ 2.8284 − 0.0894 ≈ 2.7390
Finding the Closest Value Comparing the values of f ( x ) for each option, we see that f ( − 0.50 ) ≈ 0.9670 is the closest to 1. Therefore, the solution to the equation 2 − x − 5 x = 1 to the nearest fourth of a unit is x = − 0.50 .
Examples
Imagine you are trying to balance a chemical equation where the rate of reaction depends on an exponential factor. Finding the right value of 'x' that satisfies the equation is crucial for achieving the desired reaction rate. Similarly, in financial modeling, you might use exponential equations to model the growth of an investment, and solving for 'x' helps determine the time it takes to reach a specific investment goal. This type of problem is also applicable in physics, such as determining the half-life of a radioactive substance.
