A certain radioactive isotope decays according to the formula N(t) = N_0 \cdot 2^{-t/8}, w… — ACT

A certain radioactive isotope decays according to the formula $N(t) = N_0 \cdot 2^{-t/8}$ , where $N_0$ is the initial amount of the isotope and $t$ is the time in years. If a sample initially contains $120$ grams of the isotope, how many years will it take for the amount of the isotope to decay to $15$ grams?

Answer: 24 years

Answer

The correct answer is 24 because substituting the initial value of 120 and the final value of 15 into the equation yields

15 = 120 \cdot 2^{-t/8}

. Dividing both sides by 120 gives

\frac{1}{8} = 2^{-t/8}

, which can be rewritten as

2^{-3} = 2^{-t/8}

. Setting the exponents equal to each other gives

-3 = -t/8

, and solving for

t

yields 24.

Step-by-Step Solution

Substitute the given values into the decay formula.

15 = 120 \cdot 2^{-t/8}

The initial amount

N_0

120

grams and the final amount

N(t)

15

grams.

Isolate the exponential term.

\frac{1}{8} = 2^{-t/8}

Divide both sides by

120

. Since

\frac{15}{120}

reduces to

\frac{1}{8}

, this isolates the base

2

term.

Write the fraction as a power with base 2.

2^{-3} = 2^{-t/8}

Using exponent rules,

\frac{1}{8} = \frac{1}{2^3} = 2^{-3}

Equate the exponents and solve for

t

t = 24

Since the bases are equal, the exponents must be equal, so

-3 = -\frac{t}{8}

which gives

t = 24

Key Concept

Solving exponential equations using a common base.

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