So: 960 = 120e^(3r) → 8 = e^(3r) → ln(8) = 3r → r = ln(8)/3 = 3ln(2)/3 = ln(2) ≈ <<log(2)=0.6931>>0.6931.

So: 960 = 120e^(3r) → 8 = e^(3r) → ln(8) = 3r → r = ln(8)/3 = 3ln(2)/3 = ln(2) ≈ <<log(2)=0.6931>>0.6931.

["### Solving the Exponential Equation: 960 = 120e^(3r) Step-by-Step Explained", "Understanding exponential equations is crucial in mathematics, finance, and science. One particularly useful transformation involves solving equations of the form ( a = be^{kr} ) for the variable ( r ). In this article, we’ll walk through how to solve the equation:", "[\n960 = 120e^{3r}\n]", "step-by-step, using properties of exponents and natural logarithms, and explore how this connects to fundamental logarithmic values like ( \ln(2) ).", "---", "Step 1: Isolate the exponential term", "Start by dividing both sides of the equation by 120:", "[\n\frac{960}{120} = e^{3r}\n]", "Simplifying the left-hand side gives:", "[\n8 = e^{3r}\n]", "This removes the coefficient and isolates the exponential expression ( e^{3r} ).", "---", "Step 2: Apply the natural logarithm to both sides", "To solve for ( r ), take the natural logarithm (ln) of both sides. Remember that ( \ln(e^{x}) = x ), due to the inverse relationship between ( e^x ) and ( \ln(x) ):", "[\n\ln(8) = \ln(e^{3r}) = 3r\n]", "This transforms the equation from an exponential into a linear form.", "---", "Step 3: Solve for ( r )", "Now divide both sides by 3 to isolate ( r ):", "[\nr = \frac{\ln(8)}{3}\n]", "This is the exact solution in terms of natural logarithms.", "---", "Step 4: Simplify using logarithmic identities", "We know that ( 8 = 2^3 ), so:", "[\n\ln(8) = \ln(2^3) = 3\ln(2)\n]", "Substitute this back into the expression for ( r ):", "[\nr = \frac{3\ln(2)}{3} = \ln(2)\n]", "---", "Step 5: Compute the numerical value", "Using the known approximation ( \ln(2) \approx 0.6931 ), we find:", "[\nr \approx 0.6931\n]", "This confirms that the solution to ( 960 = 120e^{3r} ) is ( r = \ln(2) ), approximately 0.693.", "---", "### Summary", "The equation ( 960 = 120e^{3r} ) simplifies elegantly through algebraic and logarithmic steps:", "[\n960 = 120e^{3r} \quad \Rightarrow \quad 8 = e^{3r} \quad \Rightarrow \quad \ln(8) = 3r \quad \Rightarrow \quad r = \frac{\ln(8)}{3} = \ln(2)\n]", "This solution highlights the power of logarithms in solving exponential relationships and links neatly to fundamental constants like ( \ln(2) ).", "For a quick reference:", "- ( \ln(8) = \ln(2^3) = 3\ln(2) )\n- ( r = \ln(2) \approx 0.6931 )", "Understanding these steps builds a strong foundation for working with exponential growth models, compound interest, decay processes, and many scientific applications.", "---", "Key Takeaways:", "- Always simplify exponential equations by isolating the exponential expression first.\n- Use logarithms—especially natural logarithm—to solve for exponents.\n- Recognize and apply logarithmic identities to simplify and compute solutions efficiently.", "---", "This method applies broadly: from solving real-world mathematical problems to modeling complex natural phenomena. Mastering these techniques enhances both computational fluency and conceptual understanding."]

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