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The problem provides the exponential function \( f(x) = ab^x \), where \( a \) and \( b \) are constants, and we are given the relation:

\[
f(n-1) = f(n) + \frac{75}{100} f(n-1)
\]

We need to find the value of \( b \).

### Step-by-Step Solution:

1. **Express \( f(n) \) and \( f(n-1) \)**:
   - From \( f(x) = ab^x \), we can express the function at specific points:
     \[
     f(n) = ab^n
     \]
     \[
     f(n-1) = ab^{n-1}
     \]

2. **Substitute these expressions into the given equation**:
   The given equation is:
   \[
   f(n-1) = f(n) + \frac{75}{100} f(n-1)
   \]
   Substituting for \( f(n) \) and \( f(n-1) \):
   \[
   ab^{n-1} = ab^n + \frac{75}{100} ab^{n-1}
   \]

3. **Simplify the equation**:
   Move all terms involving \( ab^{n-1} \) to one side:
   \[
   ab^{n-1} - \frac{75}{100} ab^{n-1} = ab^n
   \]
   Factor out \( ab^{n-1} \):
   \[
   ab^{n-1} \left( 1 - \frac{75}{100} \right) = ab^n
   \]
   Simplify the expression inside the parentheses:
   \[
   ab^{n-1} \left( \frac{25}{100} \right) = ab^n
   \]
   This simplifies to:
   \[
   \frac{1}{4} ab^{n-1} = ab^n
   \]

4. **Divide both sides by \( ab^{n-1} \)**:
   Since \( ab^{n-1} \) is nonzero, we can divide both sides by it:
   \[
   \frac{1}{4} = b
   \]

Thus, the value of \( b \) is:

\[
b = \frac{1}{4}
\]

---

Would you like more details on any part of the solution, or do you have any further questions?

Here are 5 related questions for practice:
1. How would the solution change if the constant 75/100 were different?
2. What is the general solution for any exponential function \( f(x) = ab^x \)?
3. Can you derive the value of \( a \) if additional information is provided?
4. How does changing the base \( b \) affect the growth of the exponential function?
5. How can we use logarithms to find \( x \) when given \( f(x) = k \)?

**Tip:** When working with exponential equations, simplifying by factoring can often make solving for unknown constants much easier.

The exponential function f(x) is defined by f(x) = ab^x, where a and b are positive constants. If f(n-1) = f(n) + (75/100)f(n-1), where n is a constant, what is the value of b?

Learn how to solve the exponential function f(x) = ab^x with the equation f(n-1) = f(n) + (75/100)f(n-1). Detailed step-by-step solution provided to find the value of b, covering key algebraic concepts for Grades 10-12.

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