Solving Logarithmic Equations

🪵 Solving Logarithmic Equations

Part 1 of 5 — Logs & Exponentials Are Inverses

Topics in This Part

🔑 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:

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

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

💡 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.

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) $\log_{10} \frac{1}{100} = \,?$ 3) $\log_9 3 = \,?$ (fraction is fine)

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:

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

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

⚠️ 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.

Domain Check 🎯

🪵 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.).

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.

🪵 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:

$\log_b(\text{stuff}) = c \quad\Longrightarrow\quad \text{stuff} = b^{\,c}$

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 $\log_3 x = 3$ before converting, giving $x = 3^3 = 27$.

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 $10$)

🪵 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:

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

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.
• $x = -1$: argument $\log_3(-1)$ is undefined ✗ — reject (extraneous).

$\boxed{x = 3}$

⚠️ 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 $10$)

🪵 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.

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

⚠️ 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.