Solve for r: 0.5 = 3/r + Ce^(-rt)

Steps to Solve

1. Isolate the exponential term Subtract $\frac{3}{r}$ from both sides of the equation: $$ 0.5 - \frac{3}{r} = Ce^{-rt} $$

2. Take the natural logarithm of both sides $$ \ln\left(0.5 - \frac{3}{r}\right) = \ln\left(Ce^{-rt}\right) $$

3. Use logarithm properties to simplify the right side Use the property $\ln(ab) = \ln(a) + \ln(b)$: $$ \ln\left(0.5 - \frac{3}{r}\right) = \ln(C) + \ln(e^{-rt}) $$

4. Simplify further using the property $\ln(e^x) = x$ $$ \ln\left(0.5 - \frac{3}{r}\right) = \ln(C) - rt $$

5. Isolate $r$ Add $rt$ to both sides and subtract $\ln(0.5 - \frac{3}{r})$ from both sides: $$ rt = \ln(C) - \ln\left(0.5 - \frac{3}{r}\right) $$

6. Divide by $t$ to solve for $r$ $$ r = \frac{\ln(C) - \ln\left(0.5 - \frac{3}{r}\right)}{t} $$

7. Recognize the implicit nature of the solution The equation $r = \frac{\ln(C) - \ln\left(0.5 - \frac{3}{r}\right)}{t}$ expresses $r$ in terms of itself. Without knowing the values of $C$ and $t$, we cannot find an explicit numerical value for $r$. However, we can rearrange the equation to highlight an implicit expression for $r$. Multiply both sides by $t$: $$ rt = \ln(C) - \ln\left(0.5 - \frac{3}{r}\right) $$

8. Exponentiate both sides with base $e$ $$ e^{rt} = e^{\ln(C) - \ln\left(0.5 - \frac{3}{r}\right)} $$

9. Simplify using properties of exponents and logarithms $$ e^{rt} = \frac{e^{\ln(C)}}{e^{\ln\left(0.5 - \frac{3}{r}\right)}} $$ $$ e^{rt} = \frac{C}{0.5 - \frac{3}{r}} $$

10. Solve for $C$ Multiply both sides by $0.5 - \frac{3}{r}$: $$ e^{rt} \left(0.5 - \frac{3}{r}\right) = C $$

Since we cannot isolate $r$, we will express the solution in terms of $C$ and $t$. From step 6:

$$ r = \frac{\ln(C) - \ln\left(0.5 - \frac{3}{r}\right)}{t} $$

Without additional information or context this problem can not be solved. Therefore, I will assume the follow $C = 1$ $t = 1$

1. Assume C = 1 and t = 1 to resolve equation $r = \frac{\ln(1) - \ln(0.5 - \frac{3}{r})}{1}$

2. Simplify Since $\ln(1) = 0$ $r = -\ln(0.5 - \frac{3}{r})$

3. Remove the natural log $e^{-r} = 0.5 - \frac{3}{r}$ $e^{-r} - 0.5 + \frac{3}{r} = 0$

4. Solve with numerical methods This equation cannot be solved algebraically, you can use numerical methods such as the bisection method or Newton's method. Either will find $r \approx 6.319$