To solve the equation \(3^x = 12\) using an over/under table, we need to find values of \(x\) such that \(3^x\) is greater than or less than 12.
Let's calculate \(3^x\) for a couple of values of \(x\):
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Estimate with \(x = 2\): \[ 3^2 = 9 \quad (\text{which is less than } 12) \]
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Estimate with \(x = 3\): \[ 3^3 = 27 \quad (\text{which is greater than } 12) \]
Now we can create an over/under table:
| \(x\) | \(3^x\) | Comparison to 12 | |--------|---------|-------------------| | 2 | 9 | Less than 12 | | 3 | 27 | Greater than 12 |
Since \(x = 2\) gives a value less than \(12\) and \(x = 3\) gives a value greater than \(12\), the solution lies between \(2\) and \(3\).
To narrow down further, we can try values between \(2\) and \(3\):
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Estimate with \(x = 2.5\): \[ 3^{2.5} = 3^{\frac{5}{2}} = \sqrt{27} \approx 5.196 \quad (\text{which is less than } 12) \]
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Estimate with \(x = 2.8\): \[ 3^{2.8} \approx 3^{2} \times 3^{0.8} \approx 9 \times 2.297 \approx 20.673 \quad (\text{which is greater than } 12) \]
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Estimate with \(x = 2.6\): \[ 3^{2.6} \approx 3^{2} \times 3^{0.6} \approx 9 \times 1.933 \approx 17.397 \quad (\text{which is greater than } 12) \]
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Estimate with \(x = 2.4\): \[ 3^{2.4} \approx 3^{2} \times 3^{0.4} \approx 9 \times 1.515 \approx 13.635 \quad (\text{which is greater than } 12) \]
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Estimate with \(x = 2.3\): \[ 3^{2.3} \approx 3^{2} \times 3^{0.3} \approx 9 \times 1.245 \approx 11.205 \quad (\text{which is less than } 12) \]
Now we can refine our range based on these values:
- From \(x = 2.3\) (\(11.205 < 12\)) and \(x = 2.4\) (\(13.635 > 12\)):
- Now we check with \(x = 2.35\): \[ 3^{2.35} \approx 3^{2} \times 3^{0.35} \approx 9 \times 1.284 \approx 11.556 \quad (\text{which is less than } 12) \]
Next, check with \(x = 2.375\): \[ 3^{2.375} \approx 3^2 \times 3^{0.375} \approx 9 \times 1.57 \approx 14.156 \quad (\text{which is greater than } 12) \]
At this point, we have:
- \(2.35\) (\(11.556 < 12\))
- \(2.375\) (\(14.156 > 12\))
To finalize, we can do one more check.
Check with \(x = 2.36\): \[ 3^{2.36} \approx 3^{2} \times 3^{0.36} \approx 9 \times 1.291 \approx 11.62 \quad (\text{which is less than } 12) \]
Check with \(x = 2.37\): \[ 3^{2.37} \approx 3^{2} \times 3^{0.37} \approx 9 \times 1.316 \approx 11.848 \quad (\text{which is less than } 12) \]
Check with \(x = 2.38\): \[ 3^{2.38} \approx 3^{2} \times 3^{0.38} \approx 9 \times 1.342 \approx 12.077 \quad (\text{which is greater than } 12) \]
So now we have:
- \(2.37\) (returns \(11.848 < 12\))
- \(2.38\) (returns \(12.077 > 12\))
Thus we can conclude that \(x\) is approximately \(2.37\) and \(2.38\) is too high.
Finally, based on this further narrowing, we can estimate that: \[ x \approx 2.37 \quad \text{(to two decimal places)} \]
Thus, the solution to the exponential equation is: \[ \boxed{2.37} \]
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