Question

The half-life of a specific radioactive compound is 20 years. The equation used to model this half-life is y = 5(1/2)t/20, where t represents time. Currently, there are 5 pounds of this compound 15 feet below the earth's surface. When will there only be 4 pounds of this compound left? Round your answer to the nearest tenth of a year.

Answer 1

Is this the equation, where t/20 is an exponent?

\(y = 5 \left(\dfrac{1}{2} \right)^{\frac{t}{20}} \)

Answer 2

yes i think soo

Answer 3

whats t ?

Answer 4

Let y = 4 and solve for t.

Answer 5

2.5?

Answer 6

\(4 = 5 \left(\dfrac{1}{2} \right)^{\frac{t}{20}}\)

Divide both sides by 5:

\(\dfrac{4}{5} = \left(\dfrac{1}{2} \right)^{\frac{t}{20}}\)

Now take log of both sides.

Answer 7

\(\log \dfrac{4}{5} = \log \left[ \left(\dfrac{1}{2} \right)^{\frac{t}{20}} \right]\)

Answer 8

i divide (1/2) with what 20..?

Answer 9

\(\log \dfrac{4}{5} = \dfrac{t}{20}\log \dfrac{1}{2} \)

\(\dfrac{\log 4 - \log 5}{\log1 - \log2} = \dfrac{t}{20} \)

\(\dfrac{20(\log 4 - \log 5)}{- \log2} = t \)

\(\dfrac{20(\log 4 - \log 5)}{- \log2} = t \)

\(\dfrac{20( \log 5 - 2\log 2) }{\log2} = t \)

Answer 10

soo 5-2/2 ?

Answer 11

1.5

Answer 12

t = 6.4 years

Answer 13

is this rounded ?

Answer 14

Yes, it's rounded to 1 decimal place.