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.