A scientist measures the initial amount of Carbon $-14$ in a substance to be $25$ grams. $\newline$ The relationship between $A$ , the amount of Carbon $-14$ remaining in that substance, in grams, and $t$ , the elapsed time, in years, since the initial measurement is modeled by the following equation. $\newline$ $A=25 e^{-0.00012 t}$ $\newline$ In how many years will the substance contain exactly $20$ grams $(\mathrm{g})$ of Carbon $-14$ ? $\newline$ Give an exact answer expressed as a natural logarithm. $\newline$ years

Full solution

Q. A scientist measures the initial amount of Carbon

-14

in a substance to be

25

grams.

\newline

The relationship between

A

, the amount of Carbon

-14

remaining in that substance, in grams, and

t

, the elapsed time, in years, since the initial measurement is modeled by the following equation.

\newline

A=25 e^{-0.00012 t}

\newline

In how many years will the substance contain exactly

20

grams

(\mathrm{g})

of Carbon

-14

\newline

Give an exact answer expressed as a natural logarithm.

\newline

years

Set up equation: Set up the equation with the given amount of Carbon $-14$ . $\newline$ We are given the equation $A = 25e^{-0.00012t}$ and we want to find the value of $t$ when $A = 20$ grams. $\newline$ So, we set up the equation: $20 = 25e^{-0.00012t}$ .
Divide and isolate: Divide both sides of the equation by $25$ to isolate the exponential term. $\newline$ $\frac{20}{25} = \frac{25e^{(-0.00012t)}}{25}$ $\newline$ $0.8 = e^{(-0.00012t)}$
Take natural logarithm: Take the natural logarithm of both sides to solve for $t$ . $\ln(0.8) = \ln(e^{-0.00012t})$ Using the property of logarithms that $\ln(e^x) = x$ , we get: $\ln(0.8) = -0.00012t$
Solve for t: Divide both sides by $-0.00012$ to solve for $t$ . $\newline$ $t = \frac{\ln(0.8)}{-0.00012}$
Calculate value of $\newline$ $t$ : Calculate the value of $\newline$ $t$ using the natural logarithm. $\newline$ $\newline$ $t = \frac{\ln(0.8)}{-0.00012}$ $\newline$ This is the exact answer expressed as a natural logarithm.