Problem:** A car depreciates such that its value after \( t \) years is \( V(t) = 25000(0.85)^t \). After how many years will its value drop below $10,000?

["Problem:\nA car depreciates according to the function ( V(t) = 25000(0.85)^t ), where ( t ) is the number of years since purchase. After how many years will the car’s value drop below $10,000?", "---", "Understanding Car Depreciation with Exponential Functions", "When a vehicle depreciates exponentially, its value declines by a fixed percentage each year. This is modeled mathematically by an exponential decay function of the form:", "[\nV(t) = V_0 \cdot r^t\n]", "where:\n- ( V(t) ) is the car’s value after ( t ) years,\n- ( V_0 ) is the initial value,\n- ( r ) is the depreciation rate (a number between 0 and 1),\n- ( t ) is time in years.", "In our case:\n- ( V_0 = 25000 ) dollars,\n- ( r = 0.85 ), meaning the car retains 85% of its value each year.", "---", "Setting Up the Problem", "We want to find the smallest integer ( t ) such that:", "[\nV(t) = 25000(0.85)^t < 10000\n]", "---", "Step 1: Solve the inequality", "Start by dividing both sides by 25,000:", "[\n(0.85)^t < \frac{10000}{25000} = 0.4\n]", "---", "Step 2: Apply logarithms", "Take the natural logarithm (ln) of both sides:", "[\n\ln((0.85)^t) < \ln(0.4)\n]", "Using the logarithmic identity ( \ln(a^b) = b\ln(a) ):", "[\nt \cdot \ln(0.85) < \ln(0.4)\n]", "Since ( \ln(0.85) ) is negative, dividing both sides by it reverses the inequality:", "[\nt > \frac{\ln(0.4)}{\ln(0.85)}\n]", "---", "Step 3: Calculate the numerical value", "Compute the logarithms:", "[\n\ln(0.4) \approx -0.91629, \quad \ln(0.85) \approx -0.16252\n]", "[\nt > \frac{-0.91629}{-0.16252} \approx 5.642\n]", "---", "Step 4: Interpret the result", "Since ( t ) must be a whole number of years, round up to the next integer:", "[\nt = \lceil 5.642 \rceil = 6\n]", "---", "Final Answer:\nThe car’s value will drop below $10,000 after 6 years.", "---", "Summary\nUsing an exponential decay model and logarithmic analysis, we determined that a vehicle worth $25,000 that depreciates at 15% per year will fall below $10,000 in value after 6 full years. This calculation helps car owners and financial planners anticipate asset value and plan for replacement or resale value.", "Keywords: car depreciation, exponential decay formula, value drop below $10,000, $ V(t) = 25000(0.85)^t, 6-year value calculation"]