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Pre-Calculus

Pre-calculus is an interesting area of math for students because of its multi-purpose nature. It reviews previously learned topics like trigonometry, introduces new topics like matrices and determinants, and prepares students for a formal course in calculus for the following year. Pre-calculus is sometimes referred to as Algebra 3 because it builds upon basic algebraic concepts that were learned in high school. In some educational institutions, pre-calculus is divided into separate algebra and trigonometry courses to provide a solid foundation for a calculus course.

Effective and Convenient Pre-Calculus Help Is Available Online


Many students struggle with pre-calculus and need additional assistance. Reliable online help for homework assignments is available here at 24HourAnswers.com. Our online help for homework comes in two forms: you can submit your materials to our pre-calculus specialists who will offer guidance and advice to help you arrive at the appropriate solutions. Another option is to sign up for a live online session with a knowledgeable pre-calculus tutor who will provide one-one-one instruction to help you increase your level of understanding.

We Can Help You Master the Typical Content Found in Most Pre-Calculus Courses

Typical content for a regular pre-calculus course includes:

  • functions and their graphs
  • polynomial and rational functions
  • exponential and logarithmic functions
  • trigonometry
  • systems of equations and inequalities
  • matrices and determinants
  • sequences, series, and probablility

While all these topics are important, it is fair to say that the topic of exponential and logarithmic functions is the most widespread in various fields of study, from business math to theoretical physics. As an interesting example, consider the following equation where we would like to solve for x:

ex + 2e-x = 3



A quick review of three basic logarithm rules:

product rule: ln (ab) = ln a + ln b


quotient rule: ln (a/b) = ln a - ln b


power rule: ln an = n ln a



shows that none will be helpful here. One technique for solving this equation is to multiply both sides by ex. The result is:

e2x + 2 = 3ex



After subtracting 3ex from both sides, we have:

e2x - 3ex + 2 = 0



After careful consideration of this equation, you may recognize it as a quadratic, with ex taking the place of x. We can therefore factor this just as we would any quadratic equation:

(ex-2)(ex-1) = 0



Setting each factor equal to zero, we have:

ex-2 = 0 and ex-1 = 0



Adding 2 and adding 1 to the left and right equations, respectively gives:

ex = 2 and ex = 1



Now, finally, in each case we can take the natural log of both sides:

ln ex = ln 2 and ln ex = ln 1



Remembering that ln ex = x, we have our two solutions:

x = ln 2 and x = 0



This example has served to review some of the basic properties of logarithms and to illustrate an early creative twist that was necessary for solving the given equation.

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