Solve each equation for the indicated variable. Use logarithms with the appropriate bases. y = A + B(1 - e^-Cx), for x

Verified step by step guidance

1. Start with the given equation: \(y = A + B(1 - e^{-Cx})\).
2. Isolate the exponential term by subtracting \(A\) from both sides: \(y - A = B(1 - e^{-Cx})\).
3. Divide both sides by \(B\) to get: \(\frac{y - A}{B} = 1 - e^{-Cx}\).
4. Rearrange to isolate the exponential: \(e^{-Cx} = 1 - \frac{y - A}{B}\).
5. Take the natural logarithm (ln) of both sides to solve for \(x\): \(\ln\left(e^{-Cx}\right) = \ln\left(1 - \frac{y - A}{B}\right)\), then use the property \(\ln(e^u) = u\) to write \(-Cx = \ln\left(1 - \frac{y - A}{B}\right)\), and finally solve for \(x\) by dividing both sides by \(-C\).

Key Concepts

Solving Exponential Equations

This involves isolating the exponential expression and then applying logarithms to both sides to solve for the variable in the exponent. Understanding how to manipulate equations with terms like e^(-Cx) is essential for isolating x.

Properties of Logarithms

Logarithms are the inverse operations of exponentials. Key properties such as the product, quotient, and power rules help simplify expressions and solve for variables. Knowing how to apply natural logarithms (ln) is crucial when dealing with base e.

Rearranging Formulas to Solve for a Variable

This concept involves algebraic manipulation to isolate the indicated variable on one side of the equation. It requires careful steps to maintain equality, especially when variables appear inside complex expressions like exponentials.