Math Problem Statement

Solve for the exponential function equations given the following points: (0, 2) and (1, 5), (-2, 32) and (2, 8), (1/2, 3) and (1, 1), (-2, 1/2) and (2, 2).

Solution

Let’s work on the problems provided in the image step-by-step.

The equation for the exponential function is:

f(x)=baxf(x) = b \cdot a^x

To find bb and aa, we substitute the given points into the equation and solve.


Problem 25: Points (0, 2) and (1, 5)

  1. Substituting (0,2)(0, 2): 2=ba0    b=22 = b \cdot a^0 \implies b = 2

  2. Substituting (1,5)(1, 5): 5=2a1    a=525 = 2 \cdot a^1 \implies a = \frac{5}{2}

Thus, the equation is: f(x)=2(52)xf(x) = 2 \cdot \left(\frac{5}{2}\right)^x


Problem 26: Points (2,32)(-2, 32) and (2,8)(2, 8)

  1. Substituting (2,32)(-2, 32): 32=ba2    b=32a232 = b \cdot a^{-2} \implies b = 32 \cdot a^2

  2. Substituting (2,8)(2, 8): 8=ba28 = b \cdot a^2

    Substitute b=32a2b = 32 \cdot a^2: 8=(32a2)a2    8=32a48 = (32 \cdot a^2) \cdot a^2 \implies 8 = 32 \cdot a^4 a4=832=14    a2=12    a=12a^4 = \frac{8}{32} = \frac{1}{4} \implies a^2 = \frac{1}{2} \implies a = \frac{1}{\sqrt{2}}

  3. Substitute a=12a = \frac{1}{\sqrt{2}} into b=32a2b = 32 \cdot a^2: b=3212=16b = 32 \cdot \frac{1}{2} = 16

Thus, the equation is: f(x)=16(12)xf(x) = 16 \cdot \left(\frac{1}{\sqrt{2}}\right)^x


Problem 27: Points (12,3)(\frac{1}{2}, 3) and (1,1)(1, 1)

  1. Substituting (12,3)(\frac{1}{2}, 3): 3=ba1/23 = b \cdot a^{1/2}

  2. Substituting (1,1)(1, 1): 1=ba1 = b \cdot a

    From the second equation: b=1ab = \frac{1}{a}

  3. Substitute b=1ab = \frac{1}{a} into 3=ba1/23 = b \cdot a^{1/2}: 3=1aa1/2    3=a1+1/2=a1/23 = \frac{1}{a} \cdot a^{1/2} \implies 3 = a^{-1 + 1/2} = a^{-1/2} a1/2=13    a=19a^{1/2} = \frac{1}{3} \implies a = \frac{1}{9}

  4. Substitute a=19a = \frac{1}{9} into b=1ab = \frac{1}{a}: b=9b = 9

Thus, the equation is: f(x)=9(19)xf(x) = 9 \cdot \left(\frac{1}{9}\right)^x


Problem 28: Points (2,12)(-2, \frac{1}{2}) and (2,2)(2, 2)

  1. Substituting (2,12)(-2, \frac{1}{2}): 12=ba2    b=12a2\frac{1}{2} = b \cdot a^{-2} \implies b = \frac{1}{2} \cdot a^2

  2. Substituting (2,2)(2, 2): 2=ba22 = b \cdot a^2

    Substitute b=12a2b = \frac{1}{2} \cdot a^2: 2=(12a2)a2    2=12a42 = \left(\frac{1}{2} \cdot a^2\right) \cdot a^2 \implies 2 = \frac{1}{2} \cdot a^4 a4=4    a=2a^4 = 4 \implies a = \sqrt{2}

  3. Substitute a=2a = \sqrt{2} into b=12a2b = \frac{1}{2} \cdot a^2: b=122=1b = \frac{1}{2} \cdot 2 = 1

Thus, the equation is: f(x)=1(2)x=(2)xf(x) = 1 \cdot (\sqrt{2})^x = (\sqrt{2})^x


Let me know if you have questions or would like more details! Here are related questions:

  1. How do transformations affect the graph of an exponential function?
  2. What happens to an exponential function when a>1a > 1 vs. 0<a<10 < a < 1?
  3. Can you derive exponential equations from data points in a real-world context?
  4. How does the base aa influence the growth rate of the function?
  5. What is the difference between solving exponential equations and logarithmic equations?

Tip: Always use logarithms if solving for the base or exponent becomes complex!

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Math Problem Analysis

Mathematical Concepts

Exponential Functions
Algebra
Systems of Equations
Exponents

Formulas

f(x) = b * a^x
b = 2
a = 5/2
b = 32 * a^2
a^4 = 1/4
a = 1/sqrt(2)
b = 1/2 * a^2
a = sqrt(2)

Theorems

Exponential Function Form
Solving Systems of Exponential Equations

Suitable Grade Level

Grades 9-12