upgrade
upgrade

Logarithmic Functions Rules

Study smarter with Fiveable

Get study guides, practice questions, and cheatsheets for all your subjects. Join 500,000+ students with a 96% pass rate.

Get Started

Why This Matters

Logarithmic functions are the key to unlocking equations where the variable is trapped in an exponent—and that's exactly the kind of problem Algebra 2 loves to test. You're being tested on your ability to manipulate logarithmic expressions, convert between exponential and logarithmic forms, and apply the core properties (product rule, quotient rule, power rule, change of base) to simplify and solve. These skills don't just appear on multiple-choice questions; they're essential for FRQ-style problems that ask you to show your work step-by-step.

Think of logarithm rules as a toolkit for taking apart complicated expressions and rebuilding them in simpler forms. The same properties that let you expand loga(x3y)\log_a(x^3y) into separate terms also let you condense scattered logarithms back into a single expression—and the exam will test you in both directions. Don't just memorize the formulas—know when each rule applies and why it works based on the inverse relationship between logs and exponents.


The Foundation: What Logarithms Actually Mean

Before you can apply the rules, you need rock-solid understanding of what a logarithm is. A logarithm is simply asking: "What exponent gives me this result?" Every rule flows from this definition and the inverse relationship with exponentiation.

Definition of a Logarithm

  • loga(x)=y\log_a(x) = y means ay=xa^y = x—this is the conversion you'll use constantly to rewrite between forms
  • Base restrictions: a>0a > 0 and a1a \neq 1; the argument xx must also be positive (you can't take the log of zero or a negative number)
  • Inverse relationship with exponents—logarithms and exponentials "undo" each other, which is why aloga(x)=xa^{\log_a(x)} = x

Logarithm of 1

  • loga(1)=0\log_a(1) = 0 for any valid base—because a0=1a^0 = 1 for all positive aa
  • Zero output, not zero input—students often confuse this; the argument is 1, the result is 0
  • Graph interpretation: every logarithmic function crosses the x-axis at (1,0)(1, 0)

Logarithm of the Base

  • loga(a)=1\log_a(a) = 1 for any valid base—because a1=aa^1 = a
  • Quick simplification tool—spot these in complex expressions to eliminate terms instantly
  • Appears in change of base situations—knowing loga(a)=1\log_a(a) = 1 helps verify your conversions

Compare: loga(1)=0\log_a(1) = 0 vs. loga(a)=1\log_a(a) = 1—both are "automatic simplifications" you should recognize instantly, but they answer different questions: "What power gives 1?" versus "What power gives the base itself?" If you see either pattern nested in a larger expression, simplify it first.


The Big Three: Product, Quotient, and Power Rules

These three rules are the workhorses of logarithm manipulation. They let you expand a single logarithm into multiple terms (for solving) or condense multiple logarithms into one (for simplifying). The key insight: multiplication becomes addition, division becomes subtraction, and exponents become coefficients.

Product Rule

  • loga(xy)=loga(x)+loga(y)\log_a(xy) = \log_a(x) + \log_a(y)—multiplication inside the log becomes addition outside
  • Works in reverse too—you can condense loga(x)+loga(y)\log_a(x) + \log_a(y) back into loga(xy)\log_a(xy) (same base required)
  • Why it works: if am=xa^m = x and an=ya^n = y, then am+n=xya^{m+n} = xy, so the exponents add

Quotient Rule

  • loga(xy)=loga(x)loga(y)\log_a\left(\frac{x}{y}\right) = \log_a(x) - \log_a(y)—division inside becomes subtraction outside
  • Order matters—the numerator's log comes first, then subtract the denominator's log
  • Common error: students write loga(x)loga(y)=loga(xy)\log_a(x) - \log_a(y) = \log_a(x - y)—this is wrong; subtraction of logs is division of arguments

Power Rule

  • loga(xn)=nloga(x)\log_a(x^n) = n \cdot \log_a(x)—the exponent "drops down" as a coefficient
  • Essential for solving exponential equations—when the variable is in the exponent, this rule brings it down
  • Works with any exponent—including fractions (loga(x)=12loga(x)\log_a(\sqrt{x}) = \frac{1}{2}\log_a(x)) and negatives

Compare: Product Rule vs. Power Rule—both can create coefficients, but from different sources. loga(x3)\log_a(x^3) uses the power rule to get 3loga(x)3\log_a(x), while loga(xxx)\log_a(x \cdot x \cdot x) uses the product rule to get loga(x)+loga(x)+loga(x)=3loga(x)\log_a(x) + \log_a(x) + \log_a(x) = 3\log_a(x). Same result, different reasoning—exam questions may test whether you recognize both paths.


Special Bases and Conversions

Not all logarithms use arbitrary bases. Two bases dominate real-world applications, and the change of base formula lets you convert between any of them. Calculators typically only have buttons for these two special logarithms.

Common Logarithm

  • log(x)=log10(x)\log(x) = \log_{10}(x)—when no base is written, assume base 10
  • Used in scientific scales—pH, decibels, Richter scale all use base-10 logarithms
  • Calculator button: the "LOG" key computes common logarithms

Natural Logarithm

  • ln(x)=loge(x)\ln(x) = \log_e(x) where e2.718e \approx 2.718—Euler's number is the base
  • Dominant in calculus—continuous growth, derivatives, and integrals favor natural logs
  • Calculator button: the "LN" key; remember ln(e)=1\ln(e) = 1 and ln(1)=0\ln(1) = 0

Change of Base Formula

  • loga(x)=logb(x)logb(a)\log_a(x) = \frac{\log_b(x)}{\log_b(a)}—convert any logarithm to a different base
  • Calculator strategy: use loga(x)=ln(x)ln(a)\log_a(x) = \frac{\ln(x)}{\ln(a)} or log(x)log(a)\frac{\log(x)}{\log(a)} to evaluate any base
  • Proof comes from the power rule—if you set loga(x)=y\log_a(x) = y, then ay=xa^y = x, and taking logb\log_b of both sides gives the formula

Compare: Common log vs. Natural log—both follow all the same rules (product, quotient, power), but they use different bases. On the exam, don't mix up log\log and ln\ln notation. If a problem uses ln\ln, keep everything in natural logs; if it uses log\log, stick with base 10 unless told otherwise.


The Inverse Property

This property captures the fundamental relationship between logarithms and exponentiation. They are inverse operations—each undoes the other. Recognizing this pattern lets you simplify expressions that might otherwise look intimidating.

Inverse Relationship

  • aloga(x)=xa^{\log_a(x)} = x and loga(ax)=x\log_a(a^x) = x—exponentiation and logarithms cancel when bases match
  • Think of it like multiplication and division—just as 5x5=x\frac{5 \cdot x}{5} = x, the log and exponential "cancel"
  • Solving strategy: if you have loga(expression)=number\log_a(\text{expression}) = \text{number}, exponentiate both sides with base aa to eliminate the log

Compare: aloga(x)=xa^{\log_a(x)} = x vs. loga(ax)=x\log_a(a^x) = x—these are two sides of the same coin. The first starts with an exponential and uses a log to "undo" it; the second starts with a log and uses an exponential to "undo" it. FRQs often require you to apply whichever form isolates your variable.


Quick Reference Table

ConceptKey Rules/Properties
Definition & Conversionloga(x)=yay=x\log_a(x) = y \Leftrightarrow a^y = x
Product Ruleloga(xy)=loga(x)+loga(y)\log_a(xy) = \log_a(x) + \log_a(y)
Quotient Ruleloga(xy)=loga(x)loga(y)\log_a\left(\frac{x}{y}\right) = \log_a(x) - \log_a(y)
Power Ruleloga(xn)=nloga(x)\log_a(x^n) = n\log_a(x)
Special Valuesloga(1)=0\log_a(1) = 0, loga(a)=1\log_a(a) = 1
Inverse Propertiesaloga(x)=xa^{\log_a(x)} = x, loga(ax)=x\log_a(a^x) = x
Change of Baseloga(x)=logb(x)logb(a)\log_a(x) = \frac{\log_b(x)}{\log_b(a)}
Common & Natural Logslog(x)=log10(x)\log(x) = \log_{10}(x), ln(x)=loge(x)\ln(x) = \log_e(x)

Self-Check Questions

  1. Which two properties would you use to fully expand log5(x3y)\log_5\left(\frac{x^3}{y}\right), and what is the final expanded form?

  2. A student claims that log2(8)+log2(4)=log2(12)\log_2(8) + \log_2(4) = \log_2(12). Identify their error and calculate the correct answer.

  3. Compare and contrast the product rule and the power rule: when does adding logarithms give the same result as multiplying a logarithm by a coefficient?

  4. Using the change of base formula, explain how you would evaluate log7(50)\log_7(50) on a calculator that only has LOG and LN buttons.

  5. If loga(1)=0\log_a(1) = 0 and loga(a)=1\log_a(a) = 1, what is loga(a5)+loga(1)\log_a(a^5) + \log_a(1)? Which properties justify your answer?