Question #1:How do you solve for n:
1=10,000(0.97)^n-1
Question #2: n=1 t(n)=3906.25
n=2 t(n)= 3125
n=3 t(n)= 2500
n=4 t(n)= 2000
n=5 t(n)= 1600
The rule is t(n)=2000(0.8)^n-4 Is this correct?
#1 Judging by the solution to #2, I'm assuming you meant
1 = 10,000 (0.97)^(n-1)
1/10000 = 0.96^(n-1)
-4ln10 = (n-1)ln0.97
n-1 = -4ln10/ln0.97
n = 1 - 4ln10/ln0.97
#2
0.8^0 = 1. so
n=4 t(n)=2000 fits
0.8^1 = 0.8, so t(5) = 1600 fits
I assume the rest do as well, so your solution looks good. Since there's not much formatting available her, you should write it as
t(n) = 2000(0.8)^(n-4) so there's no ambiguity.
To solve question #1, we need to isolate the variable "n" in the equation 1 = 10,000(0.97)^(n-1). Here's how you can do it:
1. Divide both sides of the equation by 10,000 to get rid of the coefficient: 1/10,000 = (0.97)^(n-1).
2. Rewrite 0.97^(n-1) as a logarithmic expression to eliminate the exponent: log base 0.97 (1/10,000) = n-1.
3. Use a calculator or logarithmic rules to evaluate log base 0.97 (1/10,000). The answer should be approximately -2.29.
4. Add 1 to both sides of the equation to isolate "n": n = -2.29 + 1.
5. Simplify the expression: n ≈ -1.29.
Therefore, the value of "n" that solves the equation is approximately -1.29.
Regarding question #2, let's verify if the given rule t(n) = 2000(0.8)^(n-4) is correct. We can substitute the provided values of n and compare them to the corresponding values of t(n) to see if they match:
When n = 1, t(n) = 2000(0.8)^(1-4) = 2000(0.8)^(-3) ≈ 2000(1.25) ≈ 2500, not 3906.25.
When n = 2, t(n) = 2000(0.8)^(2-4) = 2000(0.8)^(-2) ≈ 2000(2.5) ≈ 5000, not 3125.
When n = 3, t(n) = 2000(0.8)^(3-4) = 2000(0.8)^(-1) ≈ 2000(0.625) ≈ 1250, not 2500.
When n = 4, t(n) = 2000(0.8)^(4-4) = 2000(0.8)^0 ≈ 2000(1) ≈ 2000, which matches the provided t(n) of 2000.
When n = 5, t(n) = 2000(0.8)^(5-4) = 2000(0.8)^1 ≈ 2000(0.8) ≈ 1600, which matches the provided t(n) of 1600.
Based on the comparison, the provided rule t(n) = 2000(0.8)^(n-4) is incorrect for the given values.