Question

Irrational Exponents Quick Check
5 of 55 of 5 Items








Question
Consider the function f(x)=2x
. Which of the following is a sequence of x-values that can be used to approximate 219√
?(1 point)
Responses

4, 4.4, 4.36, 4.359, 4.3589, 4.35890
4, 4.4, 4.36, 4.359, 4.3589, 4.35890

3, 3.9, 3.87, 3.873, 3.8730, 3.87298
3, 3.9, 3.87, 3.873, 3.8730, 3.87298

4, 4.1, 4.12, 4.123, 4.1231, 4.12311
4, 4.1, 4.12, 4.123, 4.1231, 4.12311

16, 21.1, 20.53, 20.521, 20.5191, 20.51916

Answer 1

To approximate \( 2^{19} \), we need to find a sequence of \( x \)-values that approaches a point such that \( f(x) = 2^x \) is close to \( 2^{19} \).

Let's analyze the given options:

1. **4, 4.4, 4.36, 4.359, 4.3589, 4.35890**
- These values are in the range of 4, which won't yield an approximation for \( 2^{19} \).

2. **3, 3.9, 3.87, 3.873, 3.8730, 3.87298**
- These values are also in the range of around 3, which again won't be close to \( 2^{19} \).

3. **4, 4.1, 4.12, 4.123, 4.1231, 4.12311**
- While these are increasing, they still do not approach 19.

4. **16, 21.1, 20.53, 20.521, 20.5191, 20.51916**
- This is the only sequence that includes numbers close to 20, which would approximate \( 2^{19} \) (since \( 2^{19} = 524288 \) and \( 2^{20} = 1048576 \)).

Thus, the best choice for approximating \( 2^{19} \) is:

**16, 21.1, 20.53, 20.521, 20.5191, 20.51916**