Question

Which exponential equation contains the points (−1,5)
(
−
1
,
5
)
and (2,5000)
(
2
,
5000
)
?(1 point)
Responses

y=(500)x
𝑦
=
(
500
)
𝑥
y is equal to 500 to the x th power

y=50⋅−(10)x
𝑦
=
50
⋅
−
(
10
)
𝑥
y is equal to 50 times negative 10 to the x th power

y=50⋅(10)x
𝑦
=
50
⋅
(
10
)
𝑥
y is equal to 50 times 10 to the x th power

y=−50⋅(10)x

Answer 1

To determine which exponential equation contains the points \((-1, 5)\) and \((2, 5000)\), we can use the general form of an exponential function:

\[ y = a \cdot b^x \]

where \(a\) is a constant and \(b\) is the base of the exponential function.

Let's analyze each candidate equation:

1. **For \(y = 500^x\):**
- When \(x = -1\): \(y = 500^{-1} = \frac{1}{500} \neq 5\)
- When \(x = 2\): \(y = 500^2 = 250000 \neq 5000\)
- **Not a match.**

2. **For \(y = 50 \cdot (-10)^x\):**
- This equation will not yield real values for positive \(x\) since it involves negative bases raised to real numbers.
- **Not a match.**

3. **For \(y = 50 \cdot 10^x\):**
- When \(x = -1\): \(y = 50 \cdot 10^{-1} = 50 \cdot 0.1 = 5\) (Match for point (-1, 5))
- When \(x = 2\): \(y = 50 \cdot 10^2 = 50 \cdot 100 = 5000\) (Match for point (2, 5000))
- **This is a match.**

4. **For \(y = -50 \cdot (10)^x\):**
- When \(x = -1\): \(y = -50 \cdot (10)^{-1} = -50 \cdot 0.1 = -5 \neq 5\)
- When \(x = 2\): \(y = -50 \cdot (10)^2 = -50 \cdot 100 = -5000 \neq 5000\)
- **Not a match.**

Based on this analysis, the correct equation that contains both points \((-1, 5)\) and \((2, 5000)\) is:

**\(y = 50 \cdot (10)^x\)**.