There are many applications of exponentials and logarithms, including exponential growth and decay, half life, doubling time, Carbon dating, compound interest. Here are a couple of examples.

You find out that in the year 1800 an ancestor of yours invested 100 dollars at 6 percent annual interest, compounded yearly. You happen to be her sole known descendant and in the year 2005 you collect the accumulated tidy sum of dollars. You retire and devote the next 10 years of your life to writing a detailed biography of your remarkable ancestor.

Strontium-90 is a biologically important radioactive isotope that is created in nuclear explosions. It has a half life of 28 years. To reduce the amount created in a particular explosion by a factor 1,000 you would have to wait years. Round your answer to the nearest integer.

Seeds found in a grave in Egypt proved to have only 53% of the Carbon-14 of living tissue. Those seeds were harvested years ago. The half life of Carbon-14 is 5,730 years.

a) $15,406,443

b) 279.04 years = 279 years.

c) 5246.9 years = 5247 years.

Step-by-step explanation:

Compound interest

The final amount obtainable, A, from saving an initial amount, P, compounded at a rate of r in t number of years is given as

A = P (1 + r)ᵗ

A = ?

P = $100

r = 6% = 0.06

t = 2005 - 1800 = 205

A = 100 (1 + 0.06)²⁰⁵ = 100 × 154064.43 = $15406443

b) Radioactivity

Let the initial amount of Strontium be A

After 1 half life,

Amount remaining is A/2

After two half lives,

Amount remaining = A/2²

After 3 half lives,

Amount remaining = A/2³

After n half lives,

Amount remaining = A/2ⁿ

So, for this question,

(A)/(A/2ⁿ) = 1000

2ⁿ = 1000

In 2ⁿ = In 1000

n = (In 1000)/(In 2)

n = 9.966

1 half life = 28 years

n half lives = n × 28 = 9.966 × 28 = 279.04 years.

c) Carbon dating

The general relation of amount left to amount of Carbon-14 that is started with follows a first order rate of decay kinetics like every radioactive decay

A = A₀ e⁻ᵏᵗ

A = amount of Carbon-14 left at any time

A₀ = initial amount of Carbon-14

k = rate constant = (In 2)/(half life) = 0.693/5730 = 0.000121 /year.

(A/A₀) = 53% = 0.53

e⁻ᵏᵗ = 0.53

-kt = In 0.53 = -0.6349

t = 0.6349/k = 0.6349/0.000121 = 5246.9 years = 5247 years

45

step-by-step explanation:

f(x)=2/3x-8, which give g(x) the greatest value.

step-by-step explanation: