The amount of a radioactive isotope remaining after t days is modeled by the function A(t)… — SAT

The amount of a radioactive isotope remaining after $t$ days is modeled by the function $A(t) = A_0(0.64)^{\frac{t}{2}}$ , where $A_0$ is the initial amount of the isotope. If the function is rewritten in the form $A(t) = A_0(1 - r)^t$ , where $r$ is the daily decay rate, what is the value of $r$ ?

A
0.18
0.20Cevap
C
0.36
D
0.59

Cevap

The daily decay rate

r

0.20

The correct value of

r

0.20

. By rewriting the given expression using exponent rules, we obtain

(0.64)^{t/2} = ((0.64)^{1/2})^t = (0.8)^t

. Setting the base equal to the target decay rate form gives

1 - r = 0.8

, which simplifies to

r = 0.20

Adım Adım Çözüm

Apply the power of a power exponent rule,

(x^a)^b = x^{ab}

, to rewrite the exponential term.

A(t) = A_0\left(0.64^{1/2}\right)^t

This isolates the variable

t

as the exponent, allowing us to find the daily decay factor.

Evaluate the base

0.64^{1/2}

\sqrt{0.64} = 0.8

, so the expression becomes

A(t) = A_0(0.8)^t

Calculating the square root of

0.64

gives the daily decay factor of

0.8

Equate the daily decay factor

0.8

to the target form

1 - r

and solve for

r

1 - r = 0.8 \implies r = 0.20

Solving the equation gives the value of the daily decay rate.

Anahtar Kavram

Rewriting exponential equations by manipulating bases and exponents

Alternatif Yöntem

We can solve for

r

by choosing a specific value for

t

, such as

t = 2

. After 2 days, the remaining amount is

A(2) = A_0(0.64)^1 = 0.64A_0

. Using the target equation form, the remaining amount after 2 days is

A_0(1 - r)^2

. Setting the two expressions equal to each other gives

A_0(1 - r)^2 = 0.64A_0

. Dividing both sides by

A_0

gives

(1 - r)^2 = 0.64

. Taking the square root of both sides gives

1 - r = 0.8

, which yields

r = 0.20

Tahmini Süre:1m 30s

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Cevap

Adım Adım Çözüm

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