Rod says that he is thinking of a one variable exponential or logarithmic function that has the 	following characteristics:
a.	has the solutions x=1 and x=2
b. 	does not contain the digit "3"
c.	contains the number -12
What are two different equations that Rod could be thinking of? Are those the only possible 		answers?

To determine the possible equations that Rod could be thinking of, let's break down the requirements one by one:

### 1. Exponential or Logarithmic Function:
Rod is thinking of either an exponential or a logarithmic function. These are typically of the forms:
- **Exponential**: \( y = a \cdot b^x \)
- **Logarithmic**: \( y = a \cdot \log_b(x) + c \)

### 2. The function has solutions at \( x = 1 \) and \( x = 2 \):
This means that when \( x = 1 \) and \( x = 2 \), the function must equal a specific value, say \( y = 0 \) (for simplicity). For logarithmic functions, this translates to:
- \( \log_b(1) = 0 \) and \( \log_b(2) = 0 \), which isn't typical, so the function might need a specific base or transformation.
  
For exponential functions, this implies:
- \( y = a \cdot b^1 \) and \( y = a \cdot b^2 \) must satisfy some equation with \( y = 0 \).

### 3. The function does not contain the digit "3":
This restriction means that no coefficients or constants should contain the digit "3".

### 4. The function contains the number -12:
This implies that the function must evaluate to -12 at some point, meaning:
- \( f(x) = -12 \) for some \( x \).

### **Constructing Two Possible Functions:**

#### **First Possibility:**
Consider an exponential function of the form:
\[ f(x) = -12 \cdot 2^{x-1} \]
- This satisfies the condition of having solutions \( x = 1 \) and \( x = 2 \) because:
  \[
  f(1) = -12 \cdot 2^{1-1} = -12 \cdot 2^0 = -12 \cdot 1 = -12
  \]
  \[
  f(2) = -12 \cdot 2^{2-1} = -12 \cdot 2^1 = -12 \cdot 2 = -24
  \]
  
However, \( x = 1 \) is a solution that yields -12, and x = 2 provides another solution. The function does not contain the digit "3" and it contains the number -12.

#### **Second Possibility:**
Consider a logarithmic function of the form:
\[ g(x) = -12 \cdot \log_2(x) + 12 \]
- This function has solutions at \( x = 1 \) and \( x = 2 \) because:
  \[
  g(1) = -12 \cdot \log_2(1) + 12 = -12 \cdot 0 + 12 = 12
  \]
  \[
  g(2) = -12 \cdot \log_2(2) + 12 = -12 \cdot 1 + 12 = 0
  \]
  
This satisfies the condition of containing -12 because:
\[
g(x) = -12 \cdot \log_2(4) + 12 = -12 \cdot 2 + 12 = -24 + 12 = -12
\]
where \( x = 4 \). This function also does not contain the digit "3" and satisfies the number requirement.

### **Conclusion:**
These are two valid equations:
1. \( f(x) = -12 \cdot 2^{x-1} \)
2. \( g(x) = -12 \cdot \log_2(x) + 12 \)

These may not be the only possible answers, but they are correct based on the constraints provided. Other possible answers might exist, depending on transformations of logarithmic and exponential functions that meet the criteria.

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Do you want more details or have any questions?

Here are five related questions:
1. How would you solve for the x-intercepts of these functions?
2. Can the base of the logarithm be a non-integer while still satisfying these constraints?
3. What are the general properties of exponential functions compared to logarithmic ones?
4. How would you graph these functions based on the information given?
5. How would the solutions change if the functions had to include the digit "3"?

**Tip:** When analyzing a function, start by considering its basic form and transformations to see how it might meet given conditions.

Explore two different equations, an exponential function and a logarithmic function, satisfying solutions at x=1 and x=2, without containing the digit '3' and including the number -12. Discover how these functions meet these specific criteria.

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