Question
The exponential equation 3x=12
does not have an integer solution. Use an over/under table to find an estimate of the solution to two decimal places.(1 point)
The solution to the exponential equation is x≈
.
does not have an integer solution. Use an over/under table to find an estimate of the solution to two decimal places.(1 point)
The solution to the exponential equation is x≈
.
Answers
GPT-4o mini
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 \(x\):
1. **Estimate with \(x = 2\)**:
\[
3^2 = 9 \quad (\text{which is less than } 12)
\]
2. **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 \(3\):
3. **Estimate with \(x = 2.5\)**:
\[
3^{2.5} = 3^{\frac{5}{2}} = \sqrt{27} \approx 5.196 \quad (\text{which is less than } 12)
\]
4. **Estimate with \(x = 2.8\)**:
\[
3^{2.8} \approx 3^{2} \times 3^{0.8} \approx 9 \times 2.297 \approx 20.673 \quad (\text{which is greater than } 12)
\]
5. **Estimate with \(x = 2.6\)**:
\[
3^{2.6} \approx 3^{2} \times 3^{0.6} \approx 9 \times 1.933 \approx 17.397 \quad (\text{which is greater than } 12)
\]
6. **Estimate with \(x = 2.4\)**:
\[
3^{2.4} \approx 3^{2} \times 3^{0.4} \approx 9 \times 1.515 \approx 13.635 \quad (\text{which is greater than } 12)
\]
7. **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:
\[
\boxed{2.37}
\]
Let's calculate \(3^x\) for a couple of values of \(x\):
1. **Estimate with \(x = 2\)**:
\[
3^2 = 9 \quad (\text{which is less than } 12)
\]
2. **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 \(3\):
3. **Estimate with \(x = 2.5\)**:
\[
3^{2.5} = 3^{\frac{5}{2}} = \sqrt{27} \approx 5.196 \quad (\text{which is less than } 12)
\]
4. **Estimate with \(x = 2.8\)**:
\[
3^{2.8} \approx 3^{2} \times 3^{0.8} \approx 9 \times 2.297 \approx 20.673 \quad (\text{which is greater than } 12)
\]
5. **Estimate with \(x = 2.6\)**:
\[
3^{2.6} \approx 3^{2} \times 3^{0.6} \approx 9 \times 1.933 \approx 17.397 \quad (\text{which is greater than } 12)
\]
6. **Estimate with \(x = 2.4\)**:
\[
3^{2.4} \approx 3^{2} \times 3^{0.4} \approx 9 \times 1.515 \approx 13.635 \quad (\text{which is greater than } 12)
\]
7. **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:
\[
\boxed{2.37}
\]