To solve the exponential equation \( 3^x = 12 \) using an over/under table, we can evaluate \( 3^x \) for different values of \( x \) until we find two values that straddle 12.
Let's calculate a few values:
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For \( x = 2 \): \[ 3^2 = 9 \]
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For \( x = 3 \): \[ 3^3 = 27 \]
Now, we can create our table:
| \( x \) | \( 3^x \) | |----------|-----------| | 2 | 9 | | 2.5 | \( 3^{2.5} \approx 15.59 \) (since \( 3^{2.5} = 3^2 \cdot 3^{0.5} = 9 \cdot \sqrt{3} \text{ (approximately 1.732)}) \) | | 3 | 27 |
From the table, we observe the following:
- For \( x = 2 \), \( 3^2 = 9 \) (too low)
- For \( x = 2.5 \), \( 3^{2.5} \approx 15.59 \) (too high)
- For \( x = 3 \), \( 3^3 = 27 \) (too high)
Now, since \( 3^2 < 12 < 3^{2.5} \), we need to narrow it down further between 2 and 2.5.
Next, let's try \( x = 2.2 \) and \( x = 2.3 \):
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For \( x = 2.2 \): \[ 3^{2.2} \approx 9.5 \text{ (exact value calculated as } 9 \cdot 3^{0.2} = 9 \cdot 1.245 \text{ (approximately))} \]
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For \( x = 2.3 \): \[ 3^{2.3} \approx 10.7 \text{ (exact value calculated as } 9 \cdot 3^{0.3} = 9 \cdot 1.442 \text{ (approximately))} \]
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For \( x = 2.4 \): \[ 3^{2.4} \approx 12.7 \text{ (exact value calculated as } 9 \cdot 3^{0.4} = 9 \cdot 1.515 \text{ (approximately))} \]
Now we have an updated table:
| \( x \) | \( 3^x \) | |----------|-----------| | 2.2 | 9.5 | | 2.3 | 10.7 | | 2.4 | 12.7 |
Thus, \( 3^{2.3} < 12 < 3^{2.4} \).
Now we can refine further: let's try \( x = 2.35 \) and \( x = 2.38 \):
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For \( x = 2.35 \): \[ 3^{2.35} \approx 11.4 \]
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For \( x = 2.39 \): \[ 3^{2.39} \approx 12.07 \]
Now, add this to the table:
| \( x \) | \( 3^x \) | |----------|-----------| | 2.35 | 11.4 | | 2.38 | 12.07 |
Now we see that \( 3^{2.35} < 12 < 3^{2.38} \), so we refine even further between 2.37 and 2.38. Let's try:
- For \( x = 2.36 \): \[ 3^{2.36} \approx 11.76 \text{ (interpolating)} \]
At this point, since \( 3^{2.36} < 12 < 3^{2.38} \), it looks like we continue to work in this narrow range.
After estimating a few times or using a calculator, we could finally deduce that:
\[ x \approx 2.39 \]
Thus, the solution to the exponential equation \( 3^x = 12 \) is approximately:
\[ \boxed{2.40} \]