A sample of a radioactive isotope decays according to the function A(t) = A_0 \left(\frac{… — SAT

A sample of a radioactive isotope decays according to the function $A(t) = A_0 \left(\frac{1}{2}\right)^{\frac{t}{h}}$ , where $A_0$ is the initial mass of the sample, $t$ is the time in years, and $h$ is the half-life of the isotope in years. If the sample decays to $12.5\%$ of its initial mass in $45$ years, what is the value of $h$ ?

A
5.625
B
22.5
15Answer
D
135

Answer

The correct answer is 15. The remaining fraction of the radioactive isotope is

12.5\%

, which can be written as

0.125

\frac{1}{8}

. Expressing this as a power of the decay base yields

\frac{1}{8} = \left(\frac{1}{2}\right)^3

. Setting the decay formula equal to this fraction gives

\left(\frac{1}{2}\right)^{\frac{45}{h}} = \left(\frac{1}{2}\right)^3

. Equating the exponents gives

\frac{45}{h} = 3

, which yields

h = 15

Step-by-Step Solution

Set up the decay equation based on the given remaining percentage.

A(45) = 0.125 A_0

, which simplifies to

\frac{A(45)}{A_0} = 0.125 = \frac{1}{8}

This establishes the fraction of the substance remaining after

45

years.

Substitute the remaining fraction and the time

t = 45

into the exponential decay model.

\left(\frac{1}{2}\right)^{\frac{45}{h}} = \frac{1}{8}

This sets up the equation in terms of the unknown half-life parameter

h

Express both sides of the equation with a common base of

\frac{1}{2}

\left(\frac{1}{2}\right)^{\frac{45}{h}} = \left(\frac{1}{2}\right)^3

Converting

\frac{1}{8}

\left(\frac{1}{2}\right)^3

allows us to equate the exponents directly.

Equate the exponents and solve for

h

\frac{45}{h} = 3 \implies 3h = 45 \implies h = 15

Since the bases are identical, their exponents must be equal to satisfy the equation.

Key Concept

Solving exponential equations by expressing both sides with a common base.

Estimated Time:1m 30s

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