The mass of a sample of a chemical compound in a reaction decays exponentially. The mass, … — SAT

The mass of a sample of a chemical compound in a reaction decays exponentially. The mass, in grams, of the sample $t$ hours after the reaction starts can be modeled by the function $M(t) = a \cdot b^t$ , where $a$ and $b$ are positive constants. If the mass of the sample is $18$ grams after $2$ hours and $8$ grams after $4$ hours, what is the initial mass, in grams, of the sample?

Answer: 40.5

Answer

40.5

The initial mass of the sample is

40.5

grams (which can also be entered as the fraction

81/2

). This is found by setting up the two equations from the given points:

a \cdot b^2 = 18

and

a \cdot b^4 = 8

. Dividing the second equation by the first eliminates

a

and gives

b^2 = \frac{4}{9}

. Substituting

b^2 = \frac{4}{9}

back into the first equation yields

a \cdot \frac{4}{9} = 18

. Multiplying both sides by

\frac{9}{4}

results in

a = 40.5

. Since

M(0) = a \cdot b^0 = a

, the initial mass of the sample is

40.5

grams.

Step-by-Step Solution

Set up the system of exponential equations using the given coordinates.

a \cdot b^2 = 18

and

a \cdot b^4 = 8

This represents the mass of the sample at

t = 2

and

t = 4

using the model

M(t) = a \cdot b^t

Divide the second equation by the first equation to eliminate the constant

a

and solve for

b^2

b^2 = \frac{4}{9}

Dividing the equations yields

\frac{a \cdot b^4}{a \cdot b^2} = \frac{8}{18}

, which simplifies to

b^2 = \frac{4}{9}

Substitute the value of

b^2

back into the first equation to solve for the initial mass

a

a = 40.5

Substituting

b^2

gives

a \cdot \frac{4}{9} = 18

. Multiplying both sides by

\frac{9}{4}

yields

a = 18 \cdot \frac{9}{4} = 40.5

Key Concept

Solving systems of exponential equations to determine the initial value and decay factor.

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