A bug spray's effectiveness decreases over time. After h weeks, it retains only half of its
effectiveness, and this trend continues. A function that represents this is Q(t) = 30•6^5-1
A scientist testing the bug spray tried to put the given equation into the form (t) = Q0(1/2)^t/h
where t is measured in years, but she got stuck. Her work is shown here.
Step 1 Q(t) = 30•6^5-t
Step 2 Q(t) = 30•6^5 •6^-t
Step 3 Q(t) = 233,280 - (1/6)^t
Step 4 Q(1) = 233,280•(1/6)^h(t/h)
Step 5 Q(t) = 233,280 •((1/6)^h)^(t/h)
Two questions:
If (1/6)^h = 1/2, what is h to the nearest hundredth?
What is the function rewritten in the form Q(t)=Q0(1/2)^t/h
To find the value of h when (1/6)^h = 1/2, we can use logarithms.
Step 1: Start with the equation (1/6)^h = 1/2.
Step 2: Take the logarithm of both sides. Any logarithm base can be used, but let's use the natural logarithm (ln) for this example.
ln((1/6)^h) = ln(1/2)
Step 3: Use the exponent property of logarithms, which states that log(a^b) = b * log(a).
h * ln(1/6) = ln(1/2)
Step 4: Now, isolate h by dividing both sides of the equation by ln(1/6).
h = ln(1/2) / ln(1/6)
Step 5: Calculate the value of h using a calculator.
h ≈ 1.142.
So, h is approximately 1.142 when (1/6)^h = 1/2.
Now, let's rewrite the function Q(t) = 30•6^5-t in the form Q(t) = Q0(1/2)^(t/h).
Step 1: Start with the function Q(t) = 30•6^5-t.
Step 2: Substitute Q0 for 30•6^5, as Q0 represents the initial effectiveness of the bug spray.
Q(t) = Q0 • 6^-t
Step 3: Rewrite 6^-t as (1/6)^t.
Q(t) = Q0 • (1/6)^t
Step 4: Rewrite (1/6)^t as (1/2)^(t/h).
Q(t) = Q0 • (1/2)^(t/h)
So, the function rewritten in the form Q(t) = Q0(1/2)^t/h is Q(t) = Q0 • (1/2)^(t/h).
(1/6)^h = 1/2
-h log6 = -log2
h = log2/log6 = log62
or,
1/6^h = 1/2
6^h = 2
h = log62
(1/6)^t = 1/2
1/6 = (1/2)^(1/h)
(1/2)^t/h = ((1/2)^(1/h))^t = (1/6)^t