Half-life and finding a time

Exponential and logarithmic functions, grade 12 (4 units) Lesson 10 of 20Learning

Half-life and finding a time

Finding the time with a logarithm

When the time $\displaystyle t$ is in the exponent, we reach the equation $\displaystyle {q^t}={c}$ , and solve it by taking the logarithm of both sides:

t\cdot\log q=\log c\qquad\Rightarrow\qquad t=\frac{\log c}{\log q}

The first step is the law of the logarithm of a power: $\displaystyle {\log q^t}={t\cdot\log q}$ .
When $\displaystyle c$ is a power of $\displaystyle q$ ( $\displaystyle {2^t}={32}$ ) we can also compare exponents: $\displaystyle {t}={5}$ .

Dana’s savings plan

After adding 6,000 ₪, Dana’s savings plan held $\displaystyle 30333.06$ ₪, and the interest stayed 4% a year.
The example below finds after how many years the plan will hold 39,916 ₪, and after how many years the sum will triple.

$\displaystyle {30333.06\cdot1.04^t}={39916}$ , so $\displaystyle {1.04^t}={\frac{39916}{30333.06}\approx1.31592}$ .
$\displaystyle {t}={\frac{\log1.31592}{\log1.04}\approx7.00}$ : after 7 years.
The sum triples when $\displaystyle {1.04^t}={3}$ , that is $\displaystyle {t}={\frac{\log3}{\log1.04}\approx28.01}$ : after about 28 years.
To find when the sum triples we need not know the sum itself: the answer depends only on $\displaystyle q$ .

The half-life

The half-life is the time in which a decaying amount falls to a half. In every such period the amount halves again: 100 → 50 → 25 → …

\displaystyle q^{T}

\displaystyle =

\displaystyle \frac{1}{2}

$\displaystyle T$ – the half-life, in the unit of time of $\displaystyle q$ .
The half-life does not depend on the starting amount.
In the same way, in a process of growth the doubling time $\displaystyle T$ satisfies $\displaystyle {q^T}={2}$ .

A radioactive substance

Of 100 grams of radioactive substance I, 72 grams had not decayed after 4 years.
The half-life of radioactive substance II is twice the half-life of substance I.

The example below finds the half-life of substance I, and the percent by which the amount of substance II falls every year.

Substance I: $\displaystyle {100\cdot q^4}={72}$ , so $\displaystyle {q^4}={0.72}$ .
The half-life: $\displaystyle {q^T}={\frac{1}{2}}$ . Raising $\displaystyle {q^4}={0.72}$ to the power $\displaystyle \frac{T}{4}$ : $\displaystyle {0.72^{\frac{T}{4}}}={0.5}$ , so $\displaystyle {\frac{T}{4}\cdot\log0.72}={\log0.5}$ .
$\displaystyle {T}={\frac{4\log0.5}{\log0.72}\approx8.44}$ years.
Substance II: the half-life is $\displaystyle {2\cdot8.44}={16.88}$ years, so $\displaystyle {q}={0.5^{\frac{1}{16.88}}\approx0.9598}$ .
$\displaystyle {1-0.9598}={0.0402}$ : the amount of substance II falls by about 4% every year.

Comparing two processes

20,000 people live in town A, and their number grows by 5% every year.
35,000 people live in town B, and their number falls by 2% every year.

The example below finds after how many years the two towns will have the same number of residents. In the graph: $\displaystyle f$ – town A, $\displaystyle g$ – town B.

\displaystyle f(x)=20000\cdot1.05^{x}

\displaystyle g(x)=35000\cdot0.98^{x}

$\displaystyle {20000\cdot1.05^t}={35000\cdot0.98^t}$ . Dividing both sides by $\displaystyle 20000\cdot0.98^t$ : $\displaystyle {\left(\frac{1.05}{0.98}\right)^t}={1.75}$ .
$\displaystyle {t}={\frac{\log1.75}{\log\frac{1.05}{0.98}}\approx8.11}$ : after about 8 years, at the point where the graphs meet.

Exercise 1

A colony of bacteria started with 100 bacteria.
The number of bacteria doubles every hour.
Solve the equation: after how many hours will the colony have 3,200 bacteria?

Exercise 2

A sum of money was deposited in a bank at an interest of 5% a year.
After how many years will the sum double? Round to two decimal places.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power