Intermediate Algebra

• Apply exponent rules (product, quotient, power, negative, and zero exponents)
• Interpret rational exponents as roots: x^(1/n) = the nth root of x, and x^(m/n) = the nth root of x^m
• Understand logarithms as the inverse of exponents: log_b(x) = y means b^y = x
• Model exponential growth and decay: y = a(1 + r)^t for growth, y = a(1 − r)^t for decay, and continuous growth y = Pe^(rt), and solve for an exponent with a logarithm (3e^x = 12 gives x = ln 4)
• Finish on one ACT-hard question and say what made it hard

For positive real x, exponent rules give x^a × x^b = x^(a+b), x^a / x^b = x^(a−b), (x^a)^b = x^(ab), x^0 = 1, x^(−n) = 1/x^n. Zero and negative powers require a nonzero base. With negative bases, check the domain before using fractional-power rules; for example, √(x²) = |x|, not always x. For positive bases, rational (fractional) exponents mean roots: x^(1/n) is the nth root of x, and x^(m/n) is the nth root of x^m (for example 8^(2/3) = the cube root of 8, squared, = 4). Logarithms reverse exponentiation: log₂(8) = 3 because 2³ = 8. Real logarithms require positive arguments and a positive base other than 1. For positive a and b, key log rules are log(ab) = log(a) + log(b), log(a/b) = log(a) − log(b), log(a^n) = n·log(a). Exponential growth and decay: a quantity that changes by a fixed percent each period is exponential, not linear. Growth is y = a(1 + r)^t and decay is y = a(1 − r)^t, where a is the starting amount, r is the rate as a decimal, and t is the number of periods. A $500 balance growing 4% per year is 500(1.04)^t; a car losing 15% of its value per year is worth a(0.85)^t. Doubling is a(2)^t and a half-life is a(1/2)^t, with t counted in doubling periods or half-lives. Continuous growth uses the number e (about 2.718): y = Pe^(rt), so $1,000 growing continuously at 5% a year is worth 1000e^(0.05 × 10) = 1000e^0.5, about $1,649, after 10 years. The natural log, ln, is the log with base e, so it undoes e: to solve 3e^x = 12, divide by 3 first (e^x = 4), then take ln of both sides (x = ln 4); taking ln before isolating e^x is the trap. To estimate a value such as 2e^12 + 5, use e^12 ≈ 1.6 × 10^5, so the value is about 3 × 10^5. A log equation is solved by rewriting it in exponential form: y = log(5x) means 10^y = 5x, so x = 10^y/5. Four late-section topics to recognize and handle in one step (each has its own lesson for depth). Sequences: an arithmetic sequence adds the same difference d, so its nth term is a₁ + (n − 1)d; a geometric sequence multiplies by the same ratio r, so its nth term is a₁ × r^(n − 1); and 1 + 2 + ... + n = n(n + 1)/2, because pairing the first and last terms gives n/2 pairs that each add to n + 1. The same pairing adds any evenly spaced run: 5 + 8 + ... + 32 has (32 − 5)/3 + 1 = 10 terms, so its sum is 10 × (5 + 32)/2 = 185. Matrices: a matrix with m rows and n columns times one with n rows and p columns gives a matrix with m rows and p columns, and each entry is a row of the first times a column of the second. Vectors: the vector from (1, 2) to (4, 6) has components ⟨3, 4⟩ and magnitude √(3² + 4²) = 5. Powers of i: i² = −1 and the powers repeat every four (i, −1, −i, 1), so i^10 = i² = −1. These advanced topics typically appear in the later, harder questions of the Math section. Tools for the hardest exponent and log questions: to solve an exponential equation, rewrite both sides with the same base and set the exponents equal (9^x = 27 becomes 3^(2x) = 3³, so x = 3/2). A log with no base written means base 10. A quantity that doubles every k time units is a(2)^(t/k), and one that halves every k time units is a(1/2)^(t/k), because t/k counts the doublings or half-lives. After solving a log equation, reject any answer that makes a log's argument zero or negative. The hardest questions, at the end of the section: combining logs that carry coefficients (move each coefficient up as an exponent first, then combine: 2 log x − log 4 = log(x²/4), not log(2x/4)), and counting the terms of an evenly spaced run that does not start at 1.

• If you see a logarithm question and freeze, convert it to exponential form: log_b(x) = y → b^y = x.
• Negative exponents mean 'reciprocal,' not 'negative number'. This is a common misconception the ACT exploits.
• A fixed amount added each period is linear; a fixed percent each period is exponential. Decide which before writing the equation.