Algebra II: Exponential Word Problems
This is a free lesson from our course in Algebra II
 
   
This lesson explains how to solve the real-world exponential growth and decay problems. Exponential decay is generally applied to word problems that involve financial applications as well as those that deal with radioactive decay, medicine dosages, and population decline. To decay exponentially means that the topic being studied is decreasing in proportion to the amount that was previously present. This type of problem requires that you write an exponential decay function based on given information. You must then correctly substitute given values for variables and solve the equation you obtain. In solving the equation you must convert the exponential equation to a log equation and correctly use various properties of logarithms. At the conclusion of the problem, you should always check for the reasonableness of your solution. (More text below video...)
<h2> Exponential Word Problems - Watch video (Algebra II)</h2> <p> video, AlgebraII, practice questions, quizzes, subject, math help,exponential,real world problem,growth,decay,growth and decay problem,word problems,problem, example.</p> <p> how to solve the real-world exponential growth and decay problems.</p>
Other useful lessons:
Word Problems-Exponential Growth
(Continued from above) In order to solve such problems, remember to:
know how to solve an exponential equation
know the properties of logarithms
know how to analyze and understand the problem
write and solve an equation for the problem.
Exponential functions are of the form y = ax + b, where a is any real constant. It illustrates how to solve the word problems for which the half-life is given using the formula
       A = A0ekt,
where A is the amount/quantity left after t years, A0 is the initial amount, t is the number of years and k is the decay constant. Simply to say, look at exponential function x3, where 3 is the base and x is the exponent and the latter is variable. The general exponential function for x = 1.2, looks like: f(1.2) = 1.276 1.2. The formula developed from defined functions application can be used to solve related word problems.
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