To find the smallest integer $k$ that satisfies the inequality:
$$\frac{3^{1-k}}{3000} < 1$$
We can solve it step-by-step:
Step 1: Simplify the inequalityMultiply both sides by $3000$:
$$3^{1-k} < 3000$$
Step 2: Understand the relationship with $k$We are looking for the
smallest integer $k$. Because $k$ has a negative sign in the exponent ($1-k$), making $k$ smaller will make the exponent $1-k$ larger, which in turn increases the value of $3^{1-k}$. Therefore, the smallest possible value of $k$ will give us the largest power of $3$ that remains strictly less than $3000$.
Step 3: Test the powers of 3Let's find the powers of $3$ closest to $3000$:
- $3^5 = 243$
- $3^6 = 729$
- $3^7 = 2187$ (This is less than 3000)
- $3^8 = 6561$ (This exceeds 3000)
Thus, the maximum integer value that the exponent $(1-k)$ can take is $7$.
Step 4: Solve for $k$$$1 - k \le 7$$
$$1 - 7 \le k$$
$$-6 \le k$$
Since $k$ must be greater than or equal to $-6$, the smallest integer value $k$ can take is
$-6$.
Correct Answer:E) -6architkap
What is the smallest integer k such that \(\frac{3^{1-k}}{3000}<1?\)
A) -2
B) -3
C) -4
D) -5
E) -6