8 times 17 is 136
so 8 is your answer
Answer:
statement number 3
Step-by-step explanation:
modes are both 1
Section 4.6 Exponential and Logarithmic Models
While we have explored some basic applications of exponential and logarithmic functions, in this section we explore some important applications in more depth.
Radioactive Decay
In an earlier section, we discussed radioactive decay – the idea that radioactive isotopes change over time. One of the common terms associated with radioactive decay is half- life.
Half Life
The half-life of a radioactive isotope is the time it takes for half the substance to decay.
Given the basic exponential growth/decay equation h(t) = abt , half-life can be found by solving for when half the original amount remains; by solving 1 a = a(b)t , or more
2
simply 1 = bt . Notice how the initial amount is irrelevant when solving for half-life. 2
Example 1
Bismuth-210 is an isotope that decays by about 13% each day. What is the half-life of Bismuth-210?
We were not given a starting quantity, so we could either make up a value or use an unknown constant to represent the starting amount. To show that starting quantity does not affect the result, let us denote the initial quantity by the constant a. Then the decay of Bismuth-210 can be described by the equation Q(d) = a(0.87)d .
To find the half-life, we want to determine when the remaining quantity is half the original: 1a. Solving,
2
1 a = a(0.87)d 2
1 = 0.87d 2
1 d log2=log(0.87 )
Divide by a,
Take the log of both sides Usetheexponentpropertyoflogs
