A radioactive substance decays according to the equation N(t) = N_0 \left(\frac{1}{2}\righ… — SAT

A radioactive substance decays according to the equation $N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{8}}$ , where $N(t)$ is the amount of the substance remaining after $t$ days, and $N_0$ is the initial amount. If the substance decays to $\frac{1}{16}$ of its initial amount in $d$ days, what is the value of $d$ ?

32Answer
B
4
C
2
D
128

Answer

The correct answer is 32. The remaining amount of the substance is given as

\frac{1}{16}

of the initial amount, so

\frac{N(d)}{N_0} = \frac{1}{16}

. Substituting

d

for

t

in the decay formula yields

\left(\frac{1}{2}\right)^{\frac{d}{8}} = \frac{1}{16}

. Since

\frac{1}{16} = \left(\frac{1}{2}\right)^4

, we can set the exponents equal to each other:

\frac{d}{8} = 4

. Solving for

d

gives

d = 32

Step-by-Step Solution

Set up the equation for the remaining fraction of the substance.

\frac{N(d)}{N_0} = \frac{1}{16}

The problem states the substance decays to

\frac{1}{16}

of its initial amount

N_0

d

days.

Substitute the given decay formula into the equation.

\left(\frac{1}{2}\right)^{\frac{d}{8}} = \frac{1}{16}

By replacing

t

with

d

in the formula

\frac{N(t)}{N_0} = \left(\frac{1}{2}\right)^{\frac{t}{8}}

, we get the expression for the remaining fraction.

Rewrite

\frac{1}{16}

as a power with base

\frac{1}{2}

to solve for the exponent.

\left(\frac{1}{2}\right)^4 = \frac{1}{16}

Since

2^4 = 16

, the fraction

\frac{1}{16}

is equivalent to

\left(\frac{1}{2}\right)^4

Equate the exponents and solve for

d

\frac{d}{8} = 4 \implies d = 32

Since the bases are equal, their exponents must be equal. Multiplying both sides by 8 isolates

d

Key Concept

Solving exponential equations by equating bases

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