Given that the function $f(x) = a^x + b$ passes through the points $(c, 5)$ and $(2c, 161)$, we need to find the value of $b$.

Step 1: Set up the equations using the given points.

For the point $(c, 5)$: $f(c) = a^c + b = 5 \quad \text{(Equation 1)}$

For the point $(2c, 161)$: $f(2c) = a^{2c} + b = 161 \quad \text{(Equation 2)}$

Step 2: Subtract Equation 1 from Equation 2.

Subtracting Equation 1 from Equation 2 eliminates $b$: $a^{2c} + b - (a^c + b) = 161 - 5$ $a^{2c} - a^c = 156$

Step 3: Factor the equation.

Notice that the equation can be factored as: $a^c(a^c - 1) = 156$

Step 4: Express $a^c$ from Equation 1.

From Equation 1: $a^c + b = 5 \implies a^c = 5 - b$

Step 5: Substitute $a^c = 5 - b$ into the factored equation.

Substituting $a^c = 5 - b$ into $a^c(a^c - 1) = 156$: $(5 - b)((5 - b) - 1) = 156$ $(5 - b)(4 - b) = 156$

Step 6: Expand and solve the quadratic equation.

Expanding the equation: $20 - 9b + b^2 = 156$ $b^2 - 9b + 20 - 156 = 0$ $b^2 - 9b - 136 = 0$

Step 7: Solve the quadratic equation.

To solve for $b$, use the quadratic formula: $b = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(1)(-136)}}{2(1)}$ $b = \frac{9 \pm \sqrt{81 + 544}}{2}$ $b = \frac{9 \pm \sqrt{625}}{2}$ $b = \frac{9 \pm 25}{2}$

Step 8: Determine the correct value of $b$.

The two possible solutions for $b$ are: $b = \frac{9 + 25}{2} = 17 \quad \text{or} \quad b = \frac{9 - 25}{2} = -8$

Since $b$ is a positive constant, we choose $b = 17$.

Thus, the value of $b$ is $\boxed{17}$.

Would you like more details on any step, or do you have any other questions?

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Here are some related questions you might find interesting:

1. How would you find the value of $a$ given $b = 17$?
2. What would be the next step if we wanted to find the value of $c$?
3. Can the quadratic equation be solved using any other method?
4. What if $c$ were negative—how would that affect the solution?
5. How can you verify the value of $b$ using the original equations?
6. What are some real-world applications of exponential functions like $f(x)$?
7. How would the solution change if the points were $(c, 5)$ and $(3c, 161)$?
8. What if the function were $f(x) = a^x - b$ instead? How would you approach it?

Tip: When dealing with quadratic equations, always check for possible extraneous solutions, especially when constraints are given (like $b$ being positive).