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Most-Tested Intermediate Algebra on the ACT Math Test

These three concepts (quadratics, systems of equations, logarithms) are probably ones with which you’re less a bit less familiar (especially logarithms and matrices), but they come up fairly regularly on the ACT Math Test. Here’s the basics you’ll need to know to understand these types of questions.

Quadratic equations have three terms and are in the form ax² + bx + c. An example of a quadratic is x² – 5x + 6. To find the factors of this equation, we must set up our set of two parentheses: (      )(      ). The first term in both parentheses must be x, since x multiplied by x is the only way to get x². Then we look at the coefficient of the second term, -5. It’s important to include the sign in front of the integer as part of the coefficient. One of the rules of quadratic equations is that the second terms in the two factors must add together to equal the middle term’s coefficient. So we need to think of two numbers that add together to give us -5.

Already, we can think of many combinations: -6 and 1, -2 and -3, -200 and 105. So which pair is it? Now we have to look at the integer that’s the third term of the quadratic. Here it’s  + 6. Another rule of quadratic equations is that the third term of the quadratic equation will equal the product of the second terms in the two factors. So not only do we need the two numbers to add together to equal -5, but we need them to multiply together to equal + 6. Therefore the factors must be: (x – 2) (x – 3). The “roots” or the “solutions” for this quadratic would be 2 and 3.

The ACT Math Test will often present you with two or more equations with multiple variables. Remember the “n equations with n variables rule.” If there are 2 variables in an equation (for example, x and y), then there must be 2 equations that each contain those variables in order to solve. The two common ways to solve are Substitution and Combination.

Logarithms are a unique way of writing exponents. We’re used to seeing exponents in a format like y = xa. In “logs” that equation is equal to logx(y) = a. This is the most essential piece of information you’ll need to solve logarithms. You can get more practice with logarithms on Purple Math!

Let’s try a practice logarithm problem, just like the ones you might see on Test Day:

Given that logxa = 2 and logxb = 3, what is the value of logx(ab)3?

  1. 6
  2. 15
  3. 36
  4. 54
  5. 216

Here, the term we are interested in, logx(ab)3, is equivalent to 3logx(ab).

This can also be expressed as 3logxa + 3logxb, and since we know the values of logxa and logxb, we can substitute to find the answer. logx(ab)3 = 3logxa + 3logxb = 3(2) + 3(3) = 15 (Choice B).

If you don’t know these logarithmic identities, you can still solve the problem by finding values for x, a, and b that satisfy the conditions. Then, simply calculate the value of log(x)(ab)3. The easiest way to do this is to work with a base of 10, which would mean that x = 10, a = 100, and b = 1,000. We can then calculate the answer:

logx(ab)3 = log(10)(100 * 1,000)3 = log(10)(1,000,000,000,000,000) = log(10)(1015) = 15. The answer is (B).


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