= 20 + (95 − 20)e^(−10k) → 50 = 75e^(−10k) → e^(−10k) = 50/75 = 2/3.

["Understanding the Exponential Equation: 20 + (95 − 20)e^(−10k) = 50 → 75e^(−10k) = 50 → e^(−10k) = 2/3", "When tackling exponential equations in mathematics and applied science, simplifying step-by-step is key to unlocking clarity and solving for unknowns. One such equation appears commonly in fields like physics, engineering, and exponential decay modeling:", "[\n20 + (95 − 20)e^{−10k} = 50\n]", "This equation models a situation involving exponential decay or growth, where variable ( k ) represents a time factor, rate constant, or decay parameter. Let’s explore how this equation is solved mathematically and why the final result —\n[\ne^{−10k} = \frac{2}{3}\n]\n—is both elegant and useful.", "---", "### Step-by-Step Breakdown", "Step 1: Simplify the expression inside the parentheses", "Start with:\n[\n20 + (95 − 20)e^{−10k} = 50\n]", "Simplify the coefficient:\n[\n95 − 20 = 75\n]", "So the equation becomes:\n[\n20 + 75e^{−10k} = 50\n]", "Step 2: Isolate the exponential term", "Subtract 20 from both sides:\n[\n75e^{−10k} = 50 − 20\n]\n[\n75e^{−10k} = 30\n]", "Wait — hold on! There’s a small correction needed in interpretation. Earlier simplification mistakenly claimed the left side becomes ( 75e^{−10k} = 50/75 ), but actually, we should divide both sides by 75 after subtraction — let’s clarify:", "From:\n[\n20 + 75e^{−10k} = 50\n]\nSubtract 20:\n[\n75e^{−10k} = 30\n]", "Then divide both sides by 75:\n[\ne^{−10k} = \frac{30}{75} = \frac{2}{5}\n]", "Hold — this contradicts the target result ( e^{−10k} = \frac{2}{3} ). So we revisit the original equation carefully.", "---", "### Correction: Re-express the given derivation accurately", "Assuming the intended equation is:\n[\n20 + (95 − 20)e^{−10k} = 50\n]\n[\n20 + 75e^{−10k} = 50\n]", "Subtracting 20:\n[\n75e^{−10k} = 30\n]\n[\ne^{−10k} = \frac{30}{75} = \frac{2}{5}\n]", "But the problem states:\n[\ne^{−10k} = \frac{50}{75} = \frac{2}{3}\n]", "So either there’s a typo in the equation, or a misassumption. However, based on the equation you provided—\n[\n20 + (95 − 20)e^{−10k} = 50 \rightarrow 75e^{−10k} = 50 \rightarrow e^{−10k} = \frac{50}{75} = \frac{2}{3}\n]\n— this simplification only holds if the equation was mislabeled or rephrased. The correct simplification of the exact expression is:\n[\n75e^{−10k} = 50 - 20 = 30 \quad \Rightarrow \quad e^{−10k} = \frac{30}{75} = \frac{2}{5}\n]", "---", "### Clarifying the Correct Simplified Result", "To preserve mathematical accuracy, let’s define the equation clearly:", "Given:\n[\n20 + (95 − 20)e^{−10k} = 50\n]\n[\n20 + 75e^{−10k} = 50\n]\n[\n75e^{−10k} = 30\n]\n[\ne^{−10k} = \frac{30}{75} = \frac{2}{5}\n]", "Thus, the final expression\n[\ne^{−10k} = \frac{2}{5}\n]\nis mathematically correct only if the original equation summed to 30 on the right after subtracting 20.", "But if your derivation claims:", "[\n20 + 75e^{−10k} = 50 \Rightarrow 75e^{−10k} = 50 \Rightarrow e^{−10k} = \frac{2}{3}\n]", "then the equation must have been:", "[\n20 + 50e^{−10k} = 50 \Rightarrow 50e^{−10k} = 30 \Rightarrow e^{−10k} = \frac{3}{5}\n]", "Clearly, discrepancies appear. Thus, to match your stated result, the original equation likely intended:", "[\n20 + (75)e^{−10k} = 70 \Rightarrow 75e^{−10k} = 50 \Rightarrow e^{−10k} = \frac{50}{75} = \frac{2}{3}\n]", "So the correct input equation should be:\n[\n20 + 75e^{−10k} = 70\n]", "But since we are working from your stated derivation, we accept:", "[\n20 + (95−20)e^{−10k} = 50 \Rightarrow 75e^{−10k} = 30 \Rightarrow e^{−10k} = \frac{2}{5}\n]", "However, for pedagogical clarity and alignment with your target result, consider this corrected version:", "Accurate Equation Leading to ( e^{−10k} = \frac{2}{3} ):", "[\n20 + 50e^{−10k} = 70 \Rightarrow 50e^{−10k} = 50 \Rightarrow e^{−10k} = 1\n]\nNot matching.", "But if:", "[\n20 + 75e^{−10k} = 50 \Rightarrow 75e^{−10k} = 30 \Rightarrow e^{−10k} = \frac{2}{5}\n]", "The only way to get ( \frac{2}{3} ) is:", "[\n20 + 75e^{−10k} = 70 \Rightarrow 75e^{−10k} = 50 \Rightarrow e^{−10k} = \frac{50}{75} = \frac{2}{3}\n]", "---", "### Why This Form Matters", "The exponential form ( e^{−10k} = \frac{2}{3} ) appears in contexts such as:\n- Radioactive decay, where ( k ) is the decay constant and the remaining quantity halves logarithmically.\n- Cooling models, describing how temperature approaches ambient.\n- Markov processes, governing probabilities over time.", "Solving such equations lets researchers determine half-lives, relaxation times, or proportional decay rates from observed data.", "---", "### Solving for ( k )", "From:\n[\ne^{−10k} = \frac{2}{3}\n]\ntake the natural logarithm of both sides:\n[\n−10k = \ln\left(\frac{2}{3}\right)\n]\n[\nk = -\frac{1}{10} \ln\left(\frac{2}{3}\right) = \frac{1}{10} \ln\left(\frac{3}{2}\right)\n]", "This expression is used in applications requiring explicit decay rate values — such as in pharmacokinetics or astrophysics.", "---", "### Final Thoughts", "Mathematical clarity demands precise equation framing. While your derivation yields ( \frac{2}{5} ), adjusting the original equation to\n[\n20 + 75e^{−10k} = 70\n]\nleads cleanly to\n[\ne^{−10k} = \frac{2}{3}\n]\n— a cornerstone result in exponential modeling.", "Understanding steps—subtraction, isolation, division—is essential. In advanced applications, such equations model real decay, growth, and uncertainty decay. Recognizing patterns and verifying each algebra step ensures not just correctness, but deeper insight into the behavior being modeled.", "---", "Keywords: exponential decay, ( e^{−10k} ), solve exponential equation, math derivation, decay constant, logarithmic equations, natural logarithm, scientific applications, algebra review", "Meta Description: Learn how to simplify ( 20 + (95−20)e^{−10k} = 50 ) correctly, why it may yield ( e^{−10k} = \frac{2}{5} ) with precise steps, and how to derive ( k = \frac{1}{10} \ln\left(\frac{3}{2}\right) ) for decay modeling."]