Solve each equation for the indicated variable. Use logarithms with the appropriate bases. See Example 10. y = K/(1+ae^-bx), for b

Verified step by step guidance

1. Start with the given equation: \(y = \frac{K}{1 + a e^{-bx}}\).
2. Isolate the term containing the exponential by multiplying both sides by the denominator: \(y(1 + a e^{-bx}) = K\).
3. Divide both sides by \(y\) to get: \(1 + a e^{-bx} = \frac{K}{y}\).
4. Subtract 1 from both sides to isolate the exponential term: \(a e^{-bx} = \frac{K}{y} - 1\).
5. Divide both sides by \(a\): \(e^{-bx} = \frac{\frac{K}{y} - 1}{a}\), then take the natural logarithm of both sides to solve for \(b\): \(-bx = \ln\left(\frac{\frac{K}{y} - 1}{a}\right)\), and finally solve for \(b\) by dividing both sides by \(-x\).

Key Concepts

Solving for a Variable

Solving for a variable means isolating that variable on one side of the equation. This often involves algebraic manipulation such as addition, subtraction, multiplication, division, and applying inverse operations like logarithms or exponentials to both sides.

Exponential Functions and Their Properties

Exponential functions have variables in the exponent, such as e^(-bx). Understanding how to manipulate these functions, including applying inverse operations like logarithms, is essential to isolate variables within the exponent.

Logarithms and Their Use in Solving Equations

Logarithms are the inverse operations of exponentials and are used to solve equations where the variable is in the exponent. Knowing how to apply logarithms with appropriate bases allows you to rewrite exponential equations in a solvable linear form.