🎯⭐ INTERACTIVE LESSON

Solving Logarithmic Equations

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Solving Logarithmic Equations - Complete Interactive Lesson

Part 1: Logs & Exponentials Are Inverses

🪵 Solving Logarithmic Equations

Part 1 of 5 — Logs & Exponentials Are Inverses

Topics in This Part

Section
What a Logarithm Means
Switching Between Log and Exponential Form
The Domain of a Logarithm

🔑 Key Concept: A logarithm answers an exponent question: $\log_b x$ asks "to what power must I raise $b$ to get $x$ ?" Solving a log equation always comes back to that one idea.

What a Logarithm Means

The logarithm and the exponential are two ways of writing the same fact:

$\log_b x = y \quad\Longleftrightarrow\quad b^{\,y} = x$

Concept Check 🎯

Evaluate the Log 🧮

Use the exponent question to evaluate each. (Negative and fractional answers are allowed — type them like -2 or 1/2.)

1) $\log_4 64 = \,?$ 2) $\log_{10} \frac{1}{100} = \,?$

The Domain: You Can Only Log a Positive Number

Because $b^y$ is always positive when $b > 0$ , the argument of a logarithm must be positive:

$\log_b(\text{argument}) \ \text{requires}\ \text{argument} > 0$

Domain Check 🎯

Part 2: The Three Log Properties

🪵 Solving Logarithmic Equations

Part 2 of 5 — The Three Log Properties

🔑 Why this part matters: To solve a log equation you usually must first squeeze several logs into one — or pull an exponent down. Three properties do all of that work.

The Toolkit

All three come straight from the exponent rules ( $b^m b^n = b^{m+n}$ , etc.).

Part 3: Convert to Exponential Form

🪵 Solving Logarithmic Equations

Part 3 of 5 — Convert to Exponential Form

🔑 The first solving method: When one side is a single log equal to a constant, rewrite it in exponential form. The log vanishes and ordinary algebra finishes the job.

Method 1 — One Log = A Number

Get the equation into the shape $\log_b(\text{stuff}) = c$ , then convert:

$\log_{b} ($

Part 4: Combine Logs & Check for Extraneous Roots

🪵 Solving Logarithmic Equations

Part 4 of 5 — Combine Logs & Check for Extraneous Roots

🔑 The second solving method: When logs appear on both sides, condense each side to one log, then use the one-to-one property — equal logs (same base) mean equal arguments.

Method 2 — Log = Log

The one-to-one property says a log function never repeats a value:

$\log_b M = \log_b N \quad\Longrightarrow\quad M = N \quad (M, N > 0)$

Part 5: Applications, Mixed Practice & Mastery Check

🪵 Solving Logarithmic Equations

Part 5 of 5 — Applications, Mixed Practice & Mastery Check

You can now (1) read a log as an exponent question, (2) use the three properties, (3) solve by converting to exponential form, and (4) solve log = log while screening extraneous roots. Let's apply it and lock it in.

Where Logs Show Up

Logarithmic scales compress huge ranges of numbers into manageable ones.

Scale	Formula	One unit up means…
pH (chemistry)	$\text{pH} = -\log[\text{H}^+]$

Here $b$ is the base ( $b > 0,\ b \ne 1$ ), $x$ is the argument, and $y$ is the exponent the log gives back.

Read Each One Out Loud

Logarithmic form	Question it asks	Exponential form
$\log_2 8 = 3$	$2$ to what power is $8$ ?	$2^3 = 8$
$\log_5 25 = 2$	$5$ to what power is $25$ ?	$5^2 = 25$
$\log_{10} 1000 = 3$	$10$ to what power is $1000$ ?	$10^3 = 1000$
$\log_3 \frac{1}{9} = -2$	$3$ to what power is ?

💡 Two special names. $\log x$ with no base means base $10$ (the common log). $\ln x$ means base $e \approx 2.718$ (the natural log). Same rules, just a hidden base.

(fraction is fine)

You can never take the log of $0$ or of a negative number.

Expression	Allowed?	Why
$\log_2 8$	✅	$8 > 0$
$\log_2 0$	❌	no power of $2$ equals $0$
$\log_2(-4)$	❌	no power of $2$ is negative
$\log_5(x-3)$	only if $x > 3$	need $x - 3 > 0$

⚠️ Remember this now — it is the single most important idea in the whole lesson. Some equations look solved but produce a value that makes an argument $\le 0$ . That value is rejected. We'll hunt these "extraneous solutions" in Part 4.

Property	Rule	In words
Product	$\log_b(MN) = \log_b M + \log_b N$	a log of a product is a sum
Quotient	$\log_b\!\left(\dfrac{M}{N}\right) = \log_b M - \log_b N$
Power	$\log_b(M^{\,p}) = p\,\log_b M$

Two facts you'll lean on constantly:

$\log_b b = 1 \qquad\text{and}\qquad \log_b 1 = 0$

💡 The properties run both directions. To expand, break one log into many. To condense (what we need for solving), glue many logs into one.

Worked Example — Condensing

To solve equations we almost always condense the left side down to a single logarithm.

$2\log_b x + \log_b y$

Step 1 — Power rule pulls the $2$ up as an exponent: $\log_b x^2 + \log_b y$

Step 2 — Product rule merges the sum into one log: $\log_b\!\left(x^2 y\right)$

Another one — a difference becomes a quotient:

$\log_b 12 - \log_b 4 = \log_b\!\left(\frac{12}{4}\right) = \log_b 3$

⚠️ Order matters. Apply the power rule first (exponents come out), then product/quotient. Doing it backwards is the #1 algebra-of-logs error.

Concept Check 🎯

Pick the Right Property 🔽

Choose the single rule that turns the left expression into the right one.

Condense, Then Evaluate 🧮

Condense each to a single log, then evaluate the number it equals.

1) $\log_2 5 + \log_2 6 = \log_2(?)$ → enter the argument. 2) $\log_3 54 - \log_3 2 = \log_3(?)$ → enter the argument. 3) $\log_7 7 + \log_7 1 = \,?$ → enter the value.

lo g_{b} (stuff) = c ⟹ stuff = b^{c}

Worked Example: $\log_2(x + 3) = 5$

Convert to exponential form ( $\text{stuff} = b^c$ ): $x + 3 = 2^5 = 32$ $x = 29$

✅ Check the domain & the equation: argument $= 29 + 3 = 32 > 0$ ✓, and $\log_2 32 = 5$ ✓.

Worked Example: $\log(2x - 1) = 2$

No base shown means base $10$ : $2x - 1 = 10^2 = 100$ $2x = 101 \;\Rightarrow\; x = 50.5$

Worked Example with a natural log: $\ln x = 3$

Base is $e$ , so: $x = e^3 \approx 20.09$

💡 Isolate the log first. If the equation is $4 + \log_3 x = 7$ , subtract to get $\log_3 x = 3$ converting, giving .

Walk Through It 🔽

Solve $\log_4(x - 1) = 3$ one step at a time.

Concept Check 🎯

Solve It 🧮

Convert to exponential form and solve. (Decimals where needed.)

1) $\log_2(x - 4) = 6 \;\Rightarrow\; x = \,?$ 2) $\log_3 x = 4 \;\Rightarrow\; x = \,?$ 3) $\log(x) = 3 \;\Rightarrow\; x = \,?$ (base $10$ )

Worked Example: $\log_2(3x - 1) = \log_2(x + 7)$

Same base, single log each side, so set the arguments equal: $3x - 1 = x + 7 \;\Rightarrow\; 2x = 8 \;\Rightarrow\; x = 4$

✅ Check: $3(4)-1 = 11 > 0$ and $4 + 7 = 11 > 0$ . Both arguments positive — solution stands. ✓

Combine First, Then Solve

If a side has two logs, condense it before comparing.

Worked Example: $\log_3 x + \log_3(x - 2) = 1$

Step 1 — Condense (product rule): $\log_3\big(x(x-2)\big) = 1$

Step 2 — Convert to exponential form: $x(x - 2) = 3^1 = 3$

Step 3 — Solve the quadratic: $x^2 - 2x - 3 = 0 \;\Rightarrow\; (x - 3)(x + 1) = 0 \;\Rightarrow\; x = 3 \ \text{or}\ x = -1$

Step 4 — CHECK both in the original.

$x = 3$ : arguments $3 > 0$ and $3 - 2 = 1 > 0$ ✓ — keep.
: argument is ✗ — .

$\boxed{x = 3}$

⚠️ Never skip the check. Solving the quadratic can introduce a root that breaks a log's domain. The algebra "passes" but the value is not a real solution.

Why We Check 🎯

Solve and Screen 🔽

Work through $\log_2(x) + \log_2(x + 2) = 3$ .

Find the Valid Root 🧮

Solve, then enter only the valid solution (reject extraneous roots).

1) $\log_6(2x + 4) = \log_6(x + 9) \;\Rightarrow\; x = \,?$ 2) $\log_2 x + \log_2(x - 6) = 4 \;\Rightarrow\; x = \,?$ 3) $\log(x) + \log(x - 3) = 1 \;\Rightarrow\; x = \,?$ (base $10$ )

Worked Example — pH

A solution has $[\text{H}^+] = 10^{-4}$ mol/L. Find its pH. $\text{pH} = -\log(10^{-4}) = -(-4) = 4$

Worked Example — Richter

One quake is $1000$ times as intense as another ( $\frac{I}{I_0} = 1000$ ). Its magnitude difference is: $M = \log(1000) = \log(10^3) = 3 \ \text{points stronger}$

💡 Each whole step on these scales is a factor of $10$ — that's the power of a base- $10$ log.

Apply It 🧮

1) A sample has $[\text{H}^+] = 10^{-9}$ mol/L. Its pH $= -\log(10^{-9}) = \,?$ 2) Quake A is $10{,}000$ times as intense as quake B. The Richter difference $= \log(10000) = \,?$ 3) Solve for $t$ : $\ln(t) = 0 \;\Rightarrow\; t = \,?$ (recall $\log_b 1 = 0$ )

Mixed Practice 🎯

Quick Reference

Situation	Move
$\log_b(\text{stuff}) = c$	rewrite as $\text{stuff} = b^{\,c}$
$\log_b M = \log_b N$	set $M = N$ (one-to-one)
Two logs on one side	condense with product/quotient rule first
Coefficient in front	move it up with the power rule
Every time	check each candidate: every argument must be $> 0$

⚠️ The two unbreakable habits: isolate/condense to a single log before converting, and back-substitute to discard extraneous roots.

Exit Quiz ✅

Answer all three to finish the lesson.

reject (extraneous)

Solving Logarithmic Equations

Solving Logarithmic Equations - Complete Interactive Lesson

Part 1: Logs & Exponentials Are Inverses

🪵 Solving Logarithmic Equations

Topics in This Part

What a Logarithm Means

The Domain: You Can Only Log a Positive Number

Part 2: The Three Log Properties

🪵 Solving Logarithmic Equations

The Toolkit

Part 3: Convert to Exponential Form

🪵 Solving Logarithmic Equations

Method 1 — One Log = A Number

Part 4: Combine Logs & Check for Extraneous Roots

🪵 Solving Logarithmic Equations

Method 2 — Log = Log

Part 5: Applications, Mixed Practice & Mastery Check

🪵 Solving Logarithmic Equations

Where Logs Show Up

Read Each One Out Loud

Worked Example — Condensing

Worked Example: log⁡2(x+3)=5\log_2(x + 3) = 5log2​(x+3)=5

Worked Example: log⁡(2x−1)=2\log(2x - 1) = 2log(2x−1)=2

Worked Example with a natural log: ln⁡x=3\ln x = 3lnx=3

Worked Example: log⁡2(3x−1)=log⁡2(x+7)\log_2(3x - 1) = \log_2(x + 7)log2​(3x−1)=log2​(x+7)

Combine First, Then Solve

Worked Example: log⁡3x+log⁡3(x−2)=1\log_3 x + \log_3(x - 2) = 1log3​x+log3​(x−2)=1

Worked Example — pH

Worked Example — Richter

Quick Reference

Worked Example: $\log_2(x + 3) = 5$

Worked Example: $\log(2x - 1) = 2$

Worked Example with a natural log: $\ln x = 3$

Worked Example: $\log_2(3x - 1) = \log_2(x + 7)$

Worked Example: $\log_3 x + \log_3(x - 2) = 1$