Logarithm Formulas & Rules: Change of Base, Properties, Common Mistakes

A logarithm answers the question “to what exponent must a base be raised to get a given number?” Mastering the logarithm formulas and rules—especially the change-of-base formula—lets you simplify expressions, solve equations, and evaluate logs on any calculator. Below is a concise, practice-ready guide with clear examples and one compact rules table.

What a Logarithm Is (and Why It Matters)

At its core, a logarithm is an exponent. If

$b^y=x$

$b^{y} = x$ with

$b>0$

$b > 0$ and

$b\neq1$

$b \neq = 1$ , then

$\log_b x = y$

$log_{b} x = y$ . That single identity—log ↔ exponent—is the thread that ties every rule together. When you read

$\log_{10}1000=3$

$log_{10} 1000 = 3$ , you’re really seeing “base 10 raised to the power 3 equals 1000.” When you read

$\ln(e^{2.5})=2.5$

$ln (e^{2.5}) = 2.5$ , you’re seeing the natural log “undo” the exponential

$e^{2.5}$

$e^{2.5}$ .

Why logs matter in real math and science: they linearize exponential growth (like population models), compress wide-range scales (decibels in acoustics, pH in chemistry), and help solve equations where the unknown hides in the exponent. In algebra courses, they’re the natural partner to exponent rules: each logarithm rule mirrors an exponent rule because a log is an exponent. That’s why once you trust the definition above, the standard properties will feel inevitable rather than arbitrary.

A quick domain reminder: the argument of any logarithm must be positive (

$x>0$

$x > 0$ ). Bases must satisfy

$b>0$

$b > 0$ and

$b\neq1$

$b \neq = 1$ . Those conditions quietly drive many “gotchas” later, especially when solving equations and checking solutions.

Core Logarithm Rules (Properties)

These properties of logarithms convert products to sums, quotients to differences, and powers to multiples. They’re the everyday tools you’ll use to simplify or expand expressions and to solve equations.

From the definition

$b^y=x \iff \log_b x = y$

$b^{y} = x ⟺ log_{b} x = y$ , we obtain the following core rules for

$x>0,\ y>0,\ b>0,\ b\neq1$

$x > 0, y > 0, b > 0, b \neq = 1$ :

Product rule:

$\log_b(xy)=\log_b x+\log_b y$ $log_{b} (x y) = log_{b} x + log_{b} y$
(exponent on a product is a sum of exponents)
Quotient rule:

$\log_b\!\left(\frac{x}{y}\right)=\log_b x-\log_b y$ $log_{b} (y x) = log_{b} x - log_{b} y$
Power rule:

$\log_b(x^r)=r\log_b x$ $log_{b} (x^{r}) = r log_{b} x$ for any real

$r$ $r$
Root as power:

$\log_b\!\left(\sqrt[r]{x}\right)=\frac{1}{r}\log_b x$ $log_{b} (r x) = r 1 log_{b} x$
Inverse rules:

$b^{\log_b x}=x$ $b^{l o g b x} = x$ and

$\log_b(b^y)=y$ $log_{b} (b^{y}) = y$
Base conversions (preview):

$\log_b x=\dfrac{\log_k x}{\log_k b}$ $log_{b} x = log ^{k} b log ^{k} x$ for any valid base

$k$ $k$ (this is the change-of-base rule; details next)

To keep these ideas visible at a glance, here’s a compact table that pairs each rule with a plain-English cue:

Property	Formula	Quick Use
Product	$\log_b(xy)=\log_b x+\log_b y$ $log_{b} (x y) = log_{b} x + log_{b} y$	Turn multiplication inside into addition outside
Quotient	$\log_b\!\left(\frac{x}{y}\right)=\log_b x-\log_b y$ $log_{b} (y x) = log_{b} x - log_{b} y$	Turn division inside into subtraction outside
Power	$\log_b(x^r)=r\log_b x$ $log_{b} (x^{r}) = r log_{b} x$	Pull powers down to simplify or solve
Root	$\log_b(\sqrt[r]{x})=\frac{1}{r}\log_b x$ $log_{b} (r x) = r 1 log_{b} x$	Roots become fractions of logs
Inverses	$b^{\log_b x}=x,\ \log_b(b^y)=y$ $b^{l o g b x} = x, log_{b} (b^{y}) = y$	Log and exponential undo each other
Change of base	$\log_b x=\dfrac{\log_k x}{\log_k b}$ $log_{b} x = log ^{k} b log ^{k} x$	Evaluate any log using a convenient base

Worked micro-examples (properties in action):

Product:

$\log_2(8\cdot4)=\log_2 8+\log_2 4=3+2=5.$ $log_{2} (8 \cdot 4) = log_{2} 8 + log_{2} 4 = 3 + 2 = 5.$
Quotient:

$\log_3\!\left(\frac{81}{3}\right)=\log_3 81-\log_3 3=4-1=3.$ $log_{3} (3 81) = log_{3} 81 - log_{3} 3 = 4 - 1 = 3.$
Power:

$\log_5(25^3)=3\log_5 25=3\cdot2=6.$ $log_{5} (2 5^{3}) = 3 log_{5} 25 = 3 \cdot 2 = 6.$
Inverse:

$10^{\log_{10}7}=7$ $1 0^{l o g 10 7} = 7$ and

$\log_{10}(10^{1.2})=1.2.$ $log_{10} (1 0^{1.2}) = 1.2.$

These moves are the algebraic “grammar” of logs. Once fluent, you can rearrange complicated expressions into simpler ones—and, crucially, convert an unknown stuck in an exponent into a solvable linear expression.

Change-of-Base Formula (with Derivation & Calculator Tips)

The change-of-base formula lets you compute

$\log_b x$

$log_{b} x$ on any calculator—even when it doesn’t have a dedicated base

$b$

$b$ key. For any valid base

$k$

$k$ ,

$\log_b x=\frac{\log_k x}{\log_k b}.$

$log_{b} x = log ^{k} b log ^{k} x .$

Two popular choices are

$k=10$

$k = 10$ (common log) and

$k=e$

$k = e$ (natural log). Both give the same value:

$\log_b x=\frac{\log_{10}x}{\log_{10}b}=\frac{\ln x}{\ln b}.$

$log_{b} x = log ^{10} b log ^{10} x = ln b ln x .$

Derivation (short and memorable): Let

$y=\log_b x$

$y = log_{b} x$ . By definition,

$b^y=x$

$b^{y} = x$ . Apply

$\log_k$

$log_{k}$ to both sides:

$\log_k(b^y)=\log_k x \quad\Rightarrow\quad y\cdot\log_k b=\log_k x \quad\Rightarrow\quad y=\frac{\log_k x}{\log_k b}.$

$log_{k} (b^{y}) = log_{k} x \Rightarrow y \cdot log_{k} b = log_{k} x \Rightarrow y = log ^{k} b log ^{k} x .$

Because

$y=\log_b x$

$y = log_{b} x$ , we’re done.

Calculator workflow: To evaluate

$\log_7 50$

$log_{7} 50$ , enter

$\ln 50 ÷ \ln 7$

$ln 50 \div ln 7$ . On many devices, the order

$(\ln 50)/(\ln 7)$

$(ln 50) / (ln 7)$ reduces rounding error versus

$\ln(50/7)$

$ln (50/7)$ , and it keeps the meaning transparent: “the exponent on 7 that makes 50.”

Accuracy tip: When numbers are close, keep extra digits during intermediate steps. For instance,

$\log_7 50\approx\frac{3.912023005}{1.945910149}=2.009…$

$log_{7} 50 \approx 1.945910149 3.912023005 = 2.009 \dots$ . Rounding only at the end protects your final answer.

Strategic uses beyond evaluation:

Comparing magnitudes: If you need to compare

$\log_{3}(n)$ $log_{3} (n)$ and

$\log_{5}(n)$ $log_{5} (n)$ for large

$n$ $n$ , change to a common base and note that

$\log_3 n=\dfrac{\ln n}{\ln 3}$ $log_{3} n = ln 3 ln n$ and

$\log_5 n=\dfrac{\ln n}{\ln 5}$ $log_{5} n = ln 5 ln n$ . Since

$\ln 3<\ln 5$ $ln 3 < ln 5$ ,

$\log_{3} n>\log_{5} n$ $log_{3} n > log_{5} n$ for all

$n>0$ $n > 0$ .
Simplifying expressions with mixed bases: Expressions like

$\log_{2}(x)+\log_{5}(x)$ $log_{2} (x) + log_{5} (x)$ become

$\left(\frac{1}{\ln2}+\frac{1}{\ln5}\right)\ln x$ $(l n 2 1 + l n 5 1) ln x$ after changing to base

$e$ $e$ . This unifies the base and exposes a common factor.

Solving Logarithmic and Exponential Equations

Real power comes when you use these logarithm rules to solve equations. The general path is: (1) isolate the log or the exponential, (2) apply an inverse or a property, (3) solve the resulting algebra, (4) check the domain because log arguments must be positive.

A. Logarithmic equations

Combine logs first, then exponentiate.
Solve

$\log_3(x)+\log_3(x-2)=2$ $log_{3} (x) + log_{3} (x - 2) = 2$ . Product rule gives

$\log_3\big(x(x-2)\big)=2$ $log_{3} (x (x - 2)) = 2$ . Exponentiate base 3:

$x(x-2)=3^2=9\Rightarrow x^2-2x-9=0$ $x (x - 2) = 3^{2} = 9 \Rightarrow x^{2} - 2 x - 9 = 0$ . The roots are

$x=1\pm\sqrt{10}$ $x = 1 \pm 10$ . Domain requires

$x>2$ $x > 2$ , so

$x=1+\sqrt{10}\approx4.162$ $x = 1 + 10 \approx 4.162$ is valid;

$x=1-\sqrt{10}<0$ $x = 1 - 10 < 0$ is rejected.
Use the power rule to pull down exponents.
Solve

$2\log_{10}(x)=3$ $2 log_{10} (x) = 3$ . Power rule gives

$\log_{10}(x^2)=3$ $log_{10} (x^{2}) = 3$ . Then

$x^2=10^3=1000\Rightarrow x=\sqrt{1000}\approx31.622$ $x^{2} = 1 0^{3} = 1000 \Rightarrow x = 1000 \approx 31.622$ . The domain

$x>0$ $x > 0$ means

$x\approx31.622$ $x \approx 31.622$ only.
Equations with different log bases.
Solve

$\log_{2}(x)=\log_{5}(20)$ $log_{2} (x) = log_{5} (20)$ . Apply change-of-base to each side using natural logs:

$\frac{\ln x}{\ln 2}=\frac{\ln 20}{\ln 5}\Rightarrow \ln x=\ln 2\cdot \frac{\ln 20}{\ln 5}$ $l n 2 l n x = l n 5 l n 20 \Rightarrow ln x = ln 2 \cdot l n 5 l n 20$ . Exponentiate to get

$x=\exp\!\left(\ln 2\cdot \frac{\ln 20}{\ln 5}\right)=2^{\log_{5}20}.$

$x = exp (ln 2 \cdot ln 5 ln 20) = 2^{l o g 5 20} .$

You can evaluate numerically (

$x\approx 2^{1.861…}\approx3.62$

$x \approx 2^{1.861 \dots} \approx 3.62$ ) or keep the exact exponent form.

B. Exponential equations (take logs to free the variable)

Isolate the exponential and log both sides.
Solve

$5^{2x-1}=40$ $5^{2 x - 1} = 40$ . Take natural log:

$(2x-1)\ln5=\ln40\Rightarrow 2x-1=\frac{\ln40}{\ln5}\Rightarrow x=\frac{1}{2}\!\left(\frac{\ln40}{\ln5}+1\right)\approx\frac{1}{2}(1.861+1)=1.4305.$ $(2 x - 1) ln 5 = ln 40 \Rightarrow 2 x - 1 = l n 5 l n 40 \Rightarrow x = 2 1 (l n 5 l n 40 + 1) \approx 2 1 (1.861 + 1) = 1.4305.$
Different bases? No problem.
Solve

$3^x=7$ $3^{x} = 7$ . Then

$x=\dfrac{\ln7}{\ln3}\approx1.7712$ $x = ln 3 ln 7 \approx 1.7712$ . This is the quintessential change-of-base application.
Exponential expressions hiding as products.
If

$a^{x}\cdot b^{x}=c$ $a^{x} \cdot b^{x} = c$ , rewrite as

$(ab)^x=c\Rightarrow x=\dfrac{\ln c}{\ln(ab)}$ $(ab)^{x} = c \Rightarrow x = ln ( ab ) ln c$ as long as

$ab>0$ $ab > 0$ and

$ab\neq1$ $ab \neq = 1$ . Small algebraic rewrites can save time and reduce errors.

C. Modeling quick hits: In growth/decay

$P(t)=P_0e^{kt}$

$P (t) = P_{0} e^{k t}$ , the time to reach a level

$L$

$L$ is

$t=\dfrac{1}{k}\ln\!\left(\frac{L}{P_0}\right)$

$t = k 1 ln (P ^{0} L)$ . Logs convert multiplicative change into additive time steps, which is why they dominate in half-life, doubling-time, and interest-rate calculations.

Common Mistakes and How to Avoid Them

Even strong students lose points on a handful of predictable errors. Knowing them—and the quick fixes—saves you time and frustration.

1) Dropping the domain conditions.
For

$\log_b x$

$log_{b} x$ , the argument must be positive. When solving

$\log_2(x-5)=3$

$log_{2} (x - 5) = 3$ , the solution

$x=13$

$x = 13$ is fine, but if intermediate algebra offered

$x=4$

$x = 4$ you must reject it because

$x-5$

$x - 5$ would be negative. Always do a final domain check after solving.

2) Misusing the product rule on sums.
There is no rule that turns

$\log_b(x+y)$

$log_{b} (x + y)$ into

$\log_b x+\log_b y$

$log_{b} x + log_{b} y$ . Product and quotient rules apply only to multiplication and division inside the log. If you see a sum inside, factor if possible, or use numeric evaluation, but don’t split it into separate logs.

3) Pulling coefficients inside incorrectly.
The power rule says

$\log_b(x^r)=r\log_b x$

$log_{b} (x^{r}) = r log_{b} x$ . The reverse is

$\log_b x = \frac{1}{r}\log_b(x^r)$

$log_{b} x = r 1 log_{b} (x^{r})$ , not

$\log_b(rx)$

$log_{b} (r x)$ . Multiplying the argument by

$r$

$r$ is not the same as raising it to a power.

4) Forgetting base constraints.
Log bases must satisfy

$b>0, b\neq1$

$b > 0, b \neq = 1$ . Writing

$\log_{-2}x$

$log_{-2} x$ or

$\log_1 x$

$log_{1} x$ is undefined. When changing base, choose any valid

$k$

$k$ (usually

$e$

$e$ or

$10$

$10$ ) and keep it consistent across numerator and denominator.

5) Rounding too early.
Premature rounding—especially inside change-of-base—can drift answers. Keep at least 3–4 extra digits during calculations, then round the final result. For example,

$\log_7 50$

$log_{7} 50$ stabilizes if you store full precision for

$\ln 50$

$ln 50$ and

$\ln 7$

$ln 7$ before dividing.

6) Ignoring the inverse relationship.
The pair

$\log_b$

$log_{b}$ and

$b^{x}$

$b^{x}$ are inverses. If an equation contains both, consider applying one to undo the other. For instance, if you have

$b^{\log_b f(x)}$

$b^{l o g b f (x)}$ , simplify to

$f(x)$

$f (x)$ immediately; it often collapses messy expressions and reveals the solution path.

7) Confusing

$\ln$

$ln$ and

$\log_{10}$

$log_{10}$ .

$\ln x$

$ln x$ uses base

$e\approx2.71828$

$e \approx 2.71828$ and

$\log x$

$log x$ often means base

$10$

$10$ in many contexts. In higher math,

$\log$

$log$ sometimes defaults to base

$e$

$e$ . State your base or stick to explicit notation like

$\log_{10}$

$log_{10}$ and

$\ln$

$ln$ to avoid miscommunication.

Putting it all together (mini-capstone example):
Simplify and evaluate

$E=\log_3\!\left(\dfrac{27\sqrt{9}}{3}\right)$

$E = log_{3} (327 9)$ . Inside the log,

$27=3^3$

$27 = 3^{3}$ ,

$\sqrt{9}=3$

$9 = 3$ , and the denominator is

$3$

$3$ . So

$E=\log_3\!\left(\dfrac{3^3\cdot3}{3}\right)=\log_3(3^3)=3$

$E = log_{3} (33 ^{3} \cdot 3) = log_{3} (3^{3}) = 3$ . Notice how product/quotient rules plus recognizing perfect powers make the calculation nearly mental.

What a Logarithm Is (and Why It Matters)

Core Logarithm Rules (Properties)

Change-of-Base Formula (with Derivation & Calculator Tips)

Solving Logarithmic and Exponential Equations

Common Mistakes and How to Avoid Them

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