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6.6 Exponential and logarithmic equations  (Page 5/8)

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To check the result, substitute   x = 10   into   log ( 3 x − 2 ) − log ( 2 ) = log ( x + 4 ) .

log ( 3 ( 10 ) − 2 ) − log ( 2 ) = log ( ( 10 ) + 4 )              log ( 28 ) − log ( 2 ) = log ( 14 )                             log ( 28 2 ) = log ( 14 ) The solution checks .

Using the one-to-one property of logarithms to solve logarithmic equations

For any algebraic expressions   S   and   T   and any positive real number   b , where   b ≠ 1 ,

log b S = log b T     if and only if     S = T

Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.

Given an equation containing logarithms, solve it using the one-to-one property.

  1. Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation has the form   log b S = log b T .
  2. Use the one-to-one property to set the arguments equal.
  3. Solve the resulting equation,   S = T , for the unknown.

Solving an equation using the one-to-one property of logarithms

Solve   ln ( x 2 ) = ln ( 2 x + 3 ) .

                           ln ( x 2 ) = ln ( 2 x + 3 )                               x 2 = 2 x + 3 Use the one-to-one property of the logarithm .            x 2 − 2 x − 3 = 0 Get zero on one side before factoring .       ( x − 3 ) ( x + 1 ) = 0 Factor using FOIL .                      x − 3 = 0  or  x + 1 = 0 If a product is zero, one of the factors must be zero .                               x = 3  or  x = − 1 Solve for  x .
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Solve   ln ( x 2 ) = ln 1.

x = 1   or   x = − 1

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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 . [link] 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

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 0 e ln ( 0.5 ) T t A ( t ) = A 0 e ln ( 0.5 ) t T A ( t ) = A 0 ( e ln ( 0.5 ) ) t T A ( t ) = A 0 ( 1 2 ) 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
  • y   is the amount of the substance present after time   t

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?

          y = 1000 e ln ( 0.5 ) 703,800,000 t       900 = 1000 e ln ( 0.5 ) 703,800,000 t After 10% decays, 900 grams are left .        0.9 = e ln ( 0.5 ) 703,800,000 t Divide by 1000 . ln ( 0.9 ) = ln ( e ln ( 0.5 ) 703,800,000 t ) Take ln of both sides . ln ( 0.9 ) = ln ( 0.5 ) 703,800,000 t ln ( e M ) = M              t = 703,800,000 × ln ( 0.9 ) ln ( 0.5 ) years Solve for  t .              t ≈ 106,979,777 years
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Questions & Answers

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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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