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.

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**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:

## e^{x} + 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 a^{n} = n ln a

shows that none will be helpful here. One technique for solving this equation is to multiply both sides by e^{x}. The result is:

## e^{2x} + 2 = 3e^{x}

After subtracting 3e^{x} from both sides, we have:

## e^{2x} - 3e^{x} + 2 = 0

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

## (e^{x}-2)(e^{x}-1) = 0

Setting each factor equal to zero, we have:

## e^{x}-2 = 0 and e^{x}-1 = 0

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

## e^{x} = 2 and e^{x} = 1

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

## ln e^{x} = ln 2 and ln e^{x} = ln 1

Remembering that ln e^{x} = 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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