Solving Applied Problems Using Exponential and Logarithmic Equations

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}

Analysis
Ten percent of 1000 grams is 100 grams. If 100 grams decay, the amount of uranium-235 remaining is 900 grams.


Try It #13
How long will it take before twenty percent of our 1000-gram sample of uranium-235 has decayed?

Source: Rice University, https://openstax.org/books/college-algebra/pages/6-6-exponential-and-logarithmic-equations
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