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🎯⭐ INTERACTIVE LESSON

Logarithms and Their Properties

Learn step-by-step with interactive practice!

Logarithms and Their Properties - Complete Interactive Lesson

Part 1: What Is a Logarithm?

📐 Logarithms and Their Properties

Part 1 of 5 — What Is a Logarithm?


Topics in This Part

Section
Logarithms as "the exponent"
Switching between log and exponential form
Reading off simple log values

🔑 Key Concept: A logarithm answers a single question — "What exponent do I need?" The expression log⁡bx\log_b x asks: to what power must I raise the base bb to get xx?

The Definition

A logarithm is the inverse of an exponential. They say the exact same thing in two different forms:

log⁡bx=y⟺by=x\log_b x = y \quad\Longleftrightarrow\quad b^y = x

Read log⁡bx\log_b x as "log base bb of xx." The answer (yy) is the exponent.

Examples

Question in wordsLog formExponential form
"2 to what power is 8?"log⁡28=3\log_2 8 = 323=82^3 = 8
"10 to what power is 100?"log⁡10100=2\log_{10} 100 = 2102=10010^2 = 100
"5 to what power is 1?"log⁡51=0\log_5 1 = 050=15^0 = 1

🔑 Key Idea: The base of the log becomes the base of the power, and the value of the log is the exponent. If you can rewrite a log as an exponent, you can evaluate it.

⚠️ The base bb must be positive and not equal to 1, and you can only take the log of a positive number. There is no real value for log⁡2(−4)\log_2(-4) or log⁡20\log_2 0.

Concept Check 🎯

Evaluate the Log 🧮

Ask yourself: "the base to what power gives the inside?"

1) log⁡216= ?\log_2 16 = \,? 2) log⁡5125= ?\log_5 125 = \,? 3) log⁡101000= ?\log_{10} 1000 = \,?

Two Values to Memorize

Every base gives the same answer to these two, so learn them once:

log⁡b1=0(because b0=1)\log_b 1 = 0 \qquad\text{(because } b^0 = 1\text{)}

log⁡bb=1(because b1=b)\log_b b = 1 \qquad\text{(because } b^1 = b\text{)}

For example, log⁡71=0\log_7 1 = 0 and log⁡77=1\log_7 7 = 1 — no matter the base.

💡 A handy intuition: log⁡bx\log_b x tells you roughly how many times you multiply bb by itself to reach xx. You now have the one idea the whole topic rests on: a log is an exponent. In Part 2 we use the two famous bases that appear everywhere.

Match the Value 🔽

Use log⁡b1=0\log_b 1 = 0, log⁡bb=1\log_b b = 1, and the definition.

Part 2: Common Logs, Natural Logs & Inverses

📐 Logarithms and Their Properties

Part 2 of 5 — Common Logs, Natural Logs & Inverses


🔑 The Idea: Two bases are so common they get their own shorthand. And because logs and exponentials are inverses, they undo each other — a fact that powers nearly every log problem you'll solve.

The Two Famous Bases

NameWrittenMeansWhy it matters
Common loglog⁡x\log xlog⁡10x\log_{10} xbase 10, matches our number system
Natural logln⁡x\ln xlog⁡ex\log_e xbase e≈2.718e \approx 2.718, the growth constant

When you see log⁡x\log x with no base written, it means base 10. When you see ln⁡x\ln x, the base is the special number e≈2.71828e \approx 2.71828.

Examples

log⁡1000=log⁡101000=3(103=1000)\log 1000 = \log_{10} 1000 = 3 \qquad (10^3 = 1000)

ln⁡e=log⁡ee=1(e1=e)\ln e = \log_e e = 1 \qquad (e^1 = e)

ln⁡1=0(e0=1)\ln 1 = 0 \qquad (e^0 = 1)

💡 Your calculator has two log buttons: LOG (base 10) and LN (base ee). Both follow every rule in this lesson.

Concept Check 🎯

Logs and Exponentials Undo Each Other

Because log⁡bx\log_b x and bxb^x are inverse functions, applying one then the other returns you to the start. These are the inverse (cancellation) properties:

b log⁡bx=xandlog⁡b ⁣(b x)=xb^{\,\log_b x} = x \qquad\text{and}\qquad \log_b\!\left(b^{\,x}\right) = x

Examples

ExpressionSimplifies toWhy
log⁡2 ⁣(27)\log_2\!\left(2^7\right)77log base 2 undoes "2 to the"
10 log⁡410^{\,\log 4}44base-10 power undoes common log
ln⁡ ⁣(e5)\ln\!\left(e^{5}\right)55ln⁡\ln undoes ee to the power
e ln⁡9e^{\,\ln 9}99ee to the power undoes ln⁡\ln

⚠️ The base must match. log⁡2(35)\log_2(3^5) does not simplify to 55, because the log base (22) and the power base (33) differ.

Cancel Them Out 🧮

Use blog⁡bx=xb^{\log_b x} = x and log⁡b(bx)=x\log_b(b^x) = x — but only when the bases match.

1) log⁡3 ⁣(38)= ?\log_3\!\left(3^8\right) = \,? 2) ln⁡ ⁣(e6)= ?\ln\!\left(e^{6}\right) = \,? 3) 10 log⁡12= ?10^{\,\log 12} = \,?

Will They Cancel? 🔽

Pick the simplified value. Watch whether the bases match.

Part 3: The Three Big Properties

📐 Logarithms and Their Properties

Part 3 of 5 — The Three Big Properties


🔑 The Engine of the Topic: Three rules let you break apart, combine, and move logs. They come straight from the laws of exponents, because a log is an exponent.

Product, Quotient, and Power Rules

For any valid base bb and positive numbers M,NM, N:

RuleFormulaIn words
Productlog⁡b(MN)=log⁡bM+log⁡bN\log_b(MN) = \log_b M + \log_b Nmultiply inside → add logs
Quotientlog⁡b ⁣MN=log⁡bM−log⁡bN\log_b\!\dfrac{M}{N} = \log_b M - \log_b Ndivide inside → subtract logs
Powerlog⁡b ⁣(M p)=p log⁡bM\log_b\!\left(M^{\,p}\right) = p\,\log_b Mexponent inside → multiply out front

Where they come from

Exponents add when you multiply: bx⋅by=bx+yb^x \cdot b^y = b^{x+y}. Since logs are exponents, a product inside a log becomes a sum of logs. The other two rules follow the same way.

🔑 Key Idea: These rules turn multiplication into addition, division into subtraction, and powers into multiplication — they make hard logs simple.

Worked Example: Expanding

"Expand" means write a single log as a sum/difference of simpler logs.

Expand log⁡b ⁣x3yz\log_b\!\dfrac{x^3 y}{z}

log⁡bx3yz=log⁡b(x3y)−log⁡bz(quotient rule)\log_b\frac{x^3 y}{z} = \log_b(x^3 y) - \log_b z \quad\text{(quotient rule)}

=log⁡bx3+log⁡by−log⁡bz(product rule)= \log_b x^3 + \log_b y - \log_b z \quad\text{(product rule)}

=3log⁡bx+log⁡by−log⁡bz(power rule)= 3\log_b x + \log_b y - \log_b z \quad\text{(power rule)}

⚠️ Watch the signs. Everything in the denominator gets subtracted. A common error is writing +log⁡bz+\log_b z instead of −log⁡bz-\log_b z.

Concept Check 🎯

Apply One Rule 🧮

Use the property and a known value. Recall log⁡28=3\log_2 8 = 3 and log⁡525=2\log_5 25 = 2.

1) log⁡2 ⁣(84)=4⋅log⁡28= ?\log_2\!\left(8^4\right) = 4\cdot\log_2 8 = \,? 2) log⁡5 ⁣(253)=3⋅log⁡525= ?\log_5\!\left(25^3\right) = 3\cdot\log_5 25 = \,? 3) If log⁡b6=log⁡b2+log⁡b3\log_b 6 = \log_b 2 + \log_b 3 and log⁡b2=0.43\log_b 2 = 0.43, log⁡b3=0.68\log_b 3 = 0.68, then log⁡b6= ?\log_b 6 = \,?

Worked Example: Condensing

"Condense" runs the rules backward — combine several logs into one.

Condense 3log⁡bx+log⁡by−2log⁡bz3\log_b x + \log_b y - 2\log_b z

First, use the power rule to move coefficients up as exponents:

log⁡bx3+log⁡by−log⁡bz2\log_b x^3 + \log_b y - \log_b z^2

Then product rule joins the sums, quotient rule handles the subtraction:

log⁡bx3yz2\log_b\frac{x^3 y}{z^2}

💡 Strategy: When condensing, deal with coefficients first (power rule), then combine. Added terms go on top; subtracted terms go on the bottom.

Condense Step by Step 🔽

Combine log⁡b5+2log⁡bx−log⁡by\log_b 5 + 2\log_b x - \log_b y into a single log.

Part 4: Change of Base & Solving Equations

📐 Logarithms and Their Properties

Part 4 of 5 — Change of Base & Solving Equations


🔑 Big Payoff: The change-of-base formula lets your calculator evaluate any log, and the properties let you solve equations where the unknown is trapped in an exponent.

The Change-of-Base Formula

Most calculators only do base 10 (log⁡\log) and base ee (ln⁡\ln). To evaluate any other base, rewrite it:

log⁡bx=log⁡xlog⁡b=ln⁡xln⁡b\log_b x = \frac{\log x}{\log b} = \frac{\ln x}{\ln b}

You can use either base on the right (10 or ee) — the ratio comes out the same.

Example: log⁡250\log_2 50

log⁡250=log⁡50log⁡2=1.6990.301≈5.644\log_2 50 = \frac{\log 50}{\log 2} = \frac{1.699}{0.301} \approx 5.644

Check: 25=322^5 = 32 and 26=642^6 = 64, so log⁡250\log_2 50 should fall between 55 and 66. ✓

💡 The same numerator base goes on top and the same on bottom — never mix log⁡\log on top with ln⁡\ln on bottom.

Concept Check 🎯

Solving Exponential Equations

When the variable is in the exponent, take the log of both sides, then use the power rule to bring it down.

Example: 3x=203^x = 20

log⁡3x=log⁡20  ⇒  xlog⁡3=log⁡20\log 3^x = \log 20 \;\Rightarrow\; x\log 3 = \log 20

x=log⁡20log⁡3=1.3010.477≈2.727x = \frac{\log 20}{\log 3} = \frac{1.301}{0.477} \approx 2.727

Check: 32.727≈203^{2.727}\approx 20 ✓ (and it's sensibly between 32=93^2=9 and 33=273^3=27).

Solving Logarithmic Equations: log⁡2(x)=5\log_2(x) = 5

Rewrite in exponential form: x=25=32x = 2^5 = 32.

⚠️ Always check log-equation answers in the original equation — the inside of a log must stay positive, so reject any solution that makes an argument ≤0\le 0.

Solve It 🧮

1) log⁡4x=3  ⇒  x= ?\log_4 x = 3 \;\Rightarrow\; x = \,? (rewrite as a power) 2) log⁡3(x)=0  ⇒  x= ?\log_3(x) = 0 \;\Rightarrow\; x = \,? 3) 10x=1000  ⇒  x= ?10^x = 1000 \;\Rightarrow\; x = \,?

Order the Steps 🔽

You're solving 5x=405^x = 40. Choose what happens at each stage.

Part 5: Mixed Practice & Mastery Check

📐 Logarithms and Their Properties

Part 5 of 5 — Mixed Practice & Mastery Check


You can now (1) read a log as an exponent, (2) use the common/natural logs and inverse properties, (3) expand and condense with the three rules, and (4) change base and solve equations. Let's put it together.

Quick Reference

GoalKey move
Evaluate log⁡bx\log_b xask "bb to what power is xx?"
Special valueslog⁡b1=0\log_b 1 = 0,   log⁡bb=1\;\log_b b = 1
Cancel inversesblog⁡bx=xb^{\log_b x} = x,   log⁡b(bx)=x\;\log_b(b^x) = x
Product / Quotientlog⁡bMN=log⁡bM+log⁡bN\log_b MN = \log_b M + \log_b N;   log⁡bMN=log⁡bM−log⁡bN\;\log_b\frac{M}{N} = \log_b M - \log_b N
Powerlog⁡bMp=plog⁡bM\log_b M^p = p\log_b M
Change of baselog⁡bx=ln⁡xln⁡b\log_b x = \dfrac{\ln x}{\ln b}

⚠️ Remember: there is no rule that simplifies log⁡b(M+N)\log_b(M+N) — the product rule applies to multiplication inside, never to a sum. log⁡b(M+N)≠log⁡bM+log⁡bN\log_b(M+N) \ne \log_b M + \log_b N.

Spot the Mistake 🎯

Mixed Practice 🧮

1) log⁡264= ?\log_2 64 = \,? 2) ln⁡ ⁣(e10)= ?\ln\!\left(e^{10}\right) = \,? 3) Solve log⁡5x=2  ⇒  x= ?\log_5 x = 2 \;\Rightarrow\; x = \,?

Build the Single Log 🔽

Condense log⁡ba+3log⁡bc−log⁡bd\log_b a + 3\log_b c - \log_b d one step at a time.

Exit Quiz ✅

Answer all three to finish the lesson.