The exponential function Y(x)= Ce^ax satisfies Y(0)= 2 and y(1)=1 (a) find the Constants and ax C= ax= (b)= find y(2)=



Let's break down the problem step by step:

### Given:
- The exponential function: \( Y(x) = Ce^{ax} \)
- Conditions: 
  - \( Y(0) = 2 \)
  - \( Y(1) = 1 \)

### (a) Find the constants \( C \) and \( a \)

1. **Use the first condition \( Y(0) = 2 \):**

   Substituting \( x = 0 \) into the equation:
   \[
   Y(0) = C e^{a(0)} = C \cdot 1 = C
   \]
   So, \( C = 2 \).

2. **Use the second condition \( Y(1) = 1 \):**

   Substituting \( x = 1 \) into the equation:
   \[
   Y(1) = 2 e^{a(1)} = 1
   \]
   Simplifying:
   \[
   2 e^a = 1
   \]
   Divide both sides by 2:
   \[
   e^a = \frac{1}{2}
   \]
   Taking the natural logarithm of both sides:
   \[
   a = \ln\left(\frac{1}{2}\right) = -\ln(2)
   \]
   So, \( a = -\ln(2) \).

### (b) Find \( Y(2) \)

Now that we know \( C = 2 \) and \( a = -\ln(2) \), we can find \( Y(2) \):
\[
Y(2) = 2 e^{- \ln(2) \cdot 2}
\]
Simplifying the exponent:
\[
Y(2) = 2 e^{-2 \ln(2)} = 2 e^{\ln\left(\frac{1}{2^2}\right)} = 2 \cdot \frac{1}{4} = \frac{1}{2}
\]

### Final Answers:
- \( C = 2 \)
- \( a = -\ln(2) \)
- \( Y(2) = \frac{1}{2} \)

Would you like more details on any of the steps, or have any questions?

Here are some related questions for practice:
1. What would happen if \( Y(1) = 2 \)?
2. How do you find \( Y(3) \) with the same constants?
3. Can you derive a general formula for \( Y(x) \) given other conditions?
4. How does changing \( a \) affect the shape of the exponential graph?
5. What is the significance of the constant \( C \) in exponential growth?

**Tip**: When solving for exponential equations, always check if logarithmic transformations are useful to simplify equations!

The exponential function Y(x) = Ce^ax satisfies Y(0) = 2 and Y(1) = 1. (a) Find the constants C and a. (b) Find Y(2).

Find the constants C and a in the exponential equation Y(x) = Ce^ax given Y(0) = 2 and Y(1) = 1. Learn how to use logarithmic transformations and laws of exponents to solve this problem step-by-step. Suitable for high school students in Grades 10-12.

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