Exponential Equation: Find H(x) Through Two Points

by Alex Johnson 51 views

Let's dive into finding the exponential equation in the form h(x) = a(b)^x that gracefully passes through the points (-1, 1/12) and (3, 27/4). This problem combines algebra and exponential functions, offering a solid exercise in mathematical problem-solving. Our goal is to determine the values of 'a' and 'b' that define this unique exponential curve. We'll start by using the given points to create a system of equations, and then we'll solve for 'a' and 'b' using algebraic manipulation. Mastering this technique will empower you to tackle similar problems with confidence and a deeper understanding of exponential functions.

Setting Up the Equations

Our mission is to pinpoint the exponential equation in the form h(x) = a(b)^x that gracefully traverses through the coordinates (-1, 1/12) and (3, 27/4). To achieve this, we'll use the coordinates as x and h(x) values to establish a system of equations. This system will pave the way for us to solve for the unknowns a and b, which define the specific exponential function we're seeking. By meticulously substituting these points, we transform the abstract problem into a tangible algebraic challenge, setting the stage for a methodical solution.

  1. Using the point (-1, 1/12):

    Substituting x = -1 and h(x) = 1/12 into the equation gives us:

    1/12 = a(b)^(-1) which can be rewritten as 1/12 = a/b.

  2. Using the point (3, 27/4):

    Substituting x = 3 and h(x) = 27/4 into the equation gives us:

    27/4 = a(b)^(3).

Now we have a system of two equations:

  1. a/b = 1/12
  2. a(b)^(3) = 27/4

Solving for a and b

Now that we've artfully set up our system of equations, it's time to put on our algebraic hats and solve for a and b. Our strategy will involve isolating one variable in terms of the other and then substituting that expression into the second equation. This technique, known as substitution, allows us to reduce the two equations into a single equation with a single unknown, making it solvable. Once we find the value of one variable, we can easily back-substitute to find the value of the other. This systematic approach transforms a seemingly complex problem into a series of manageable steps, showcasing the power of algebraic manipulation.

From the first equation, we can express a in terms of b:

a = b/12

Substitute this expression for a into the second equation:

(b/12) * b^(3) = 27/4

Simplify the equation:

b^(4) / 12 = 27/4

Multiply both sides by 12:

b^(4) = (27/4) * 12 = 27 * 3 = 81

Now, take the fourth root of both sides:

b = 3 (Since we are looking for exponential functions with positive bases)

Now that we have the value of b, we can find the value of a:

a = b/12 = 3/12 = 1/4

The Exponential Equation

Having meticulously solved for both a and b, we now stand at the threshold of unveiling the exponential equation that satisfies the given conditions. By substituting the values we've found, a = 1/4 and b = 3, into the general form h(x) = a(b)^x, we obtain the specific equation that passes through the points (-1, 1/12) and (3, 27/4). This final step not only provides the solution but also reinforces the connection between algebraic manipulation and the properties of exponential functions. The resulting equation represents the unique curve that precisely fits the specified coordinates, showcasing the elegance and precision of mathematical problem-solving.

Therefore, the exponential equation is:

h(x) = (1/4)(3)^x

Verification

Before we confidently declare our solution, it's crucial to verify that our derived exponential equation, h(x) = (1/4)(3)^x, indeed passes through the given points (-1, 1/12) and (3, 27/4). By substituting the x-values of these points into our equation and confirming that the resulting h(x) values match the given y-values, we can ensure the accuracy of our solution. This step serves as a final check, reinforcing our understanding of exponential functions and solidifying the correctness of our algebraic manipulations. This verification process not only validates our answer but also instills confidence in our problem-solving abilities.

  1. For the point (-1, 1/12):

    h(-1) = (1/4)(3)^(-1) = (1/4)(1/3) = 1/12

  2. For the point (3, 27/4):

    h(3) = (1/4)(3)^(3) = (1/4)(27) = 27/4

Both points satisfy the equation. Hence, our solution is correct.

Conclusion

In conclusion, we successfully determined the exponential equation h(x) = (1/4)(3)^x that elegantly passes through the points (-1, 1/12) and (3, 27/4). This journey involved setting up a system of equations, strategically solving for a and b through substitution, and rigorously verifying our solution. This exercise not only reinforced our understanding of exponential functions but also honed our algebraic problem-solving skills. Mastering these techniques empowers us to confidently tackle similar mathematical challenges and appreciate the beauty of mathematical relationships.

For further exploration of exponential functions, you can visit Khan Academy's Exponential Functions Section.