# Thread: Solving an exponential equation for t.

1. ## Solving an exponential equation for t.

Carbon-14 is a radioactive isotope of carbon which has a half-life of about 5600 years. Let R be the nearly constant ratio of corbon-12 found in the atmosphere. Let r be the ratio of carbon-14 to carbon-12 found in an observed specimen. It has been shown for carbon-14 dating of objects that
R = re (tIn4/5600) Where t is the age of the object in years. Suppose a specimen has been found in which
r = 0.2R, find the age of the specimen

2. Hello, blenyo!

Carbon-14 is a radioactive isotope of carbon which has a half-life of about 5600 years.
Let $\displaystyle R$ be the nearly constant ratio of corbon-12 found in the atmosphere.
Let $\displaystyle r$ be the ratio of carbon-14 to carbon-12 found in an observed specimen.
It has been shown for carbon-14 dating of objects that: .R = re^[(t ln4)/5600]
. . where $\displaystyle t$ is the age of the object in years.

Suppose a specimen has been found in which r = 0.2R
Find the age of the specimen.

We have: .R/r = e^[(t ln4)/5600]

We are given: .r = 0.2R . . R/r = 5

Hence: .e^[(t ln4)/5600] .= .5

. a . . . . . . .(t ln4)/5600 .= .ln5

. n . . . . . . . . . . . . . . t .= .(5600 ln5)/(ln4)

Therefore: .t . .6501.4 years.

3. By the way, the natural logarithm is ln(x), not In(x)!