Solving Applied Problems Using Exponential and Logarithmic Equations

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Solving Applied Problems Using Exponential and Logarithmic Equations

In our last section on techniques for solving exponential and logarithmic equations, you will solve applied problems involving half-life and radioactive decay of chemical elements.

In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.

One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its half-life. Table 1 lists the half-life for several of the more common radioactive substances.

Substance	Use	Half-life
gallium-67	nuclear medicine	80 hours
cobalt-60	manufacturing	5.3 years
technetium-99m	nuclear medicine	6 hours
americium-241	construction	432 years
carbon-14	archeological dating	5,715 years
uranium-235	atomic power	703,800,000 years

Table 1

We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:

$A(t)=A_0e^{\dfrac{ln(0.5)}{T}t}$

$A(t)=A_0e^{ln(0.5)\dfrac{t}{T}}$

$A(t)=A_0(e^{ln(0.5)})\dfrac{t}{T}$

$A(t)=A_0(\dfrac{1}{2})^{\dfrac{t}{T}}$

where

$A_0$ is the amount initially present
$T$ is the half-life of the substance
$t$ is the time period over which the substance is studied
$A(t)$ is the amount of the substance present after time $t$

Example 13

Using the Formula for Radioactive Decay to Find the Quantity of a Substance

How long will it take for ten percent of a 1000-gram sample of uranium-235 to decay?

Solution

$y=1000e\dfrac{ln(0.5)}{703,800,000}t$
$900=1000e^{\dfrac{ln(0.5)}{703,800,000}t}$	After 10% decays, 900 grams are left.
$0.9=e^{\dfrac{ln(0.5)}{703,800,000}t}$	Divide by 1000.
$ln(0.9)=ln(e^{\dfrac{ln(0.5)}{703,800,000}t})$	Take $ln$ of both sides.
$ln(0.9)=\dfrac{ln(0.5)}{703,800,000}t$	$ln(e^M)=M$
$t=703,800,000 × \dfrac{ln(0.9)}{ln(0.5)} \text{ years}$	Solve fort.
$t≈106,979,777\text{ years}$