To solve the equation \(200 + 50x = 1.8^x\), we can start by manipulating the equation into a more solvable form. This type of equation involves both a linear term and an exponential term, so we may need to use numerical methods or graphical methods to find an approximation of the solution.
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Rearrange the equation: \[ 1.8^x - 50x - 200 = 0 \]
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Since this equation doesn't have a straightforward algebraic solution, we can use numerical methods (like the Newton-Raphson method) or graphing to find the intersection point.
To approximate the solution, let's evaluate the left-hand side at some points:
- For \(x = 0\): \[ 1.8^0 - 50(0) - 200 = 1 - 0 - 200 = -199 \]
- For \(x = 5\): \[ 1.8^5 - 50(5) - 200 \approx 18.89 - 250 - 200 \approx -431.11 \]
- For \(x = 10\): \[ 1.8^{10} - 50(10) - 200 \approx 1073.74 - 500 - 200 = 373.74 \]
From the calculations, we see that the function changes sign between \(x = 5\) and \(x = 10\). We can now narrow it down further.
Let's check \(x = 7\) and \(x = 8\):
- For \(x = 7\): \[ 1.8^7 - 50(7) - 200 \approx 218.7 - 350 - 200 = -331.3 \]
- For \(x = 8\): \[ 1.8^8 - 50(8) - 200 \approx 392.4 - 400 - 200 = -207.6 \]
- For \(x = 9\): \[ 1.8^9 - 50(9) - 200 \approx 705.71 - 450 - 200 = 55.71 \]
From this, we can see the solution lies between \(x = 8\) and \(x = 9\). A more precise method (like interval halving or using graphing) can help find the solution to a certain level of accuracy.
Using a numerical approximation method (like Newton's method) or a graphing calculator would yield a more precise value.
If needed, I can show you how to proceed with a specific method to get an approximate numerical solution. Would you like to see that?