A bug spray's effectiveness decreases over time. After h weeks, it retains only half of its

Question

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

Answer 1

(1/6)^h = 1/2
-h log6 = -log2
h = log2/log6 = log₆2

or,
1/6^h = 1/2
6^h = 2
h = log₆2

Answer 2

(1/6)^t = 1/2
1/6 = (1/2)^(1/h)
(1/2)^t/h = ((1/2)^(1/h))^t = (1/6)^t