Log Formulas: All Logarithm Rules & Examples

#Math Formula
TL;DR
A logarithm answers one question: what power do I raise the base to in order to get this number? Every log formula on this page comes from rewriting that question.
BT
Bhanzu TeamLast updated on May 1, 202610 min read

What is a logarithm?

The log formula in math is a rewritten exponent. The expression log⁡ba=c\log_b a = c means exactly the same thing as bc=ab^c = a. The base bb is the number being raised; the answer cc is the power; aa is the result. Every log formula listed below is a way of moving between these two forms.

In school work, two bases come up most often. Base 10 is called the common logarithm and is written log⁡x\log x with no subscript. Base ee (where e≈2.71828e \approx 2.71828) is called the natural logarithm and is written ln⁡x\ln x. The rules below apply to every base.

All Log Formulas in One Table

Name

Formula

When you'd use it

Product rule

log⁡b(MN)=log⁡bM+log⁡bN\log_b(MN) = \log_b M + \log_b N

Splitting a log of a product into a sum

Quotient rule

log⁡b!(MN)=log⁡bM−log⁡bN\log_b!\left(\dfrac{M}{N}\right) = \log_b M - \log_b N

Splitting a log of a fraction into a difference

Power rule

log⁡b(Mk)=k⋅log⁡bM\log_b(M^k) = k \cdot \log_b M

Pulling an exponent out front

Zero rule

log⁡b1=0\log_b 1 = 0

b0=1b^0 = 1 for any base

Identity rule

log⁡bb=1\log_b b = 1

b1=bb^1 = b

Inverse rule

blog⁡bx=xb^{\log_b x} = x

Cancelling a base with its log

Change of base

log⁡bx=log⁡axlog⁡ab\log_b x = \dfrac{\log_a x}{\log_a b}

Computing log⁡bx\log_b x on a calculator that only has log⁡\log or ln⁡\ln

Reciprocal

log⁡ba=1log⁡ab\log_b a = \dfrac{1}{\log_a b}

Flipping the base and the argument

Negative log

log⁡b!(1N)=−log⁡bN\log_b!\left(\dfrac{1}{N}\right) = -\log_b N

Special case of the quotient rule

Equal logs imply equal arguments

If log⁡bM=log⁡bN\log_b M = \log_b N, then M=NM = N

Solving log equations

Variable Key

Symbol

Meaning

bb

Base of the logarithm. Must be positive and not equal to 1.

MM, NN

Arguments of the logarithm. Must be positive.

kk

Any real number (the exponent in the power rule).

log⁡x\log x

log⁡10x\log_{10} x — common logarithm

ln⁡x\ln x

log⁡ex\log_e x — natural logarithm

Two restrictions decide whether a log expression even exists. The base bb must be positive and must not be 1 (a base of 1 makes the log undefined because 1c=11^c = 1 for every cc). The argument must be strictly positive — there is no real-number value for log⁡b0\log_b 0 or log⁡b(−2)\log_b(-2).

When to Use Each Rule

The product, quotient, and power rules are tools for expanding a single complicated log into a sum of simpler logs, or condensing a sum of simple logs back into one. Most algebra problems involving logs need one of these moves. Change of base is for the moment your calculator has log⁡\log and ln⁡\ln but you need log⁡5\log_5 or log⁡2\log_2. The inverse rule appears whenever you solve an equation where the variable is stuck inside a log or inside an exponent.

Worked Examples of Log Formula

Example 1: Expand log⁡2(8x3)\log_2(8x^3) using the log rules.

Apply the product rule:

log⁡2(8x3)=log⁡28+log⁡2x3\log_2(8x^3) = \log_2 8 + \log_2 x^3

Apply the power rule on the second term:

=log⁡28+3log⁡2x= \log_2 8 + 3\log_2 x

Evaluate log⁡28\log_2 8 (because 23=82^3 = 8):

=3+3log⁡2x= 3 + 3\log_2 x

Final answer: log⁡2(8x3)=3+3log⁡2x\log_2(8x^3) = 3 + 3\log_2 x.

Example 2: Condense 2log⁡5x−log⁡5y2\log_5 x - \log_5 y into a single log.

Apply the power rule in reverse on the first term:

2log⁡5x=log⁡5x22\log_5 x = \log_5 x^2

Substitute back:

log⁡5x2−log⁡5y\log_5 x^2 - \log_5 y

Apply the quotient rule:

=log⁡5!(x2y)= \log_5!\left(\dfrac{x^2}{y}\right)

Final answer: 2log⁡5x−log⁡5y=log⁡5!(x2y)2\log_5 x - \log_5 y = \log_5!\left(\dfrac{x^2}{y}\right).

Example 3: Evaluate log⁡512\log_5 12 using change of base.

A calculator gives log⁡\log (base 10) and ln⁡\ln (base ee), not log⁡5\log_5. Use change of base:

log⁡512=log⁡12log⁡5\log_5 12 = \dfrac{\log 12}{\log 5}

Compute each:

log⁡12≈1.0792\log 12 \approx 1.0792

log⁡5≈0.6990\log 5 \approx 0.6990

Divide:

log⁡512≈1.07920.6990≈1.5440\log_5 12 \approx \dfrac{1.0792}{0.6990} \approx 1.5440

Final answer: log⁡512≈1.5440\log_5 12 \approx 1.5440.

Example 4: Solve log⁡3(x+2)−log⁡3x=2\log_3(x+2) - \log_3 x = 2.

Combine the left side using the quotient rule:

log⁡3!(x+2x)=2\log_3!\left(\dfrac{x+2}{x}\right) = 2

Rewrite in exponential form (this is the inverse rule in action):

x+2x=32=9\dfrac{x+2}{x} = 3^2 = 9

Solve the linear equation:

x+2=9xx + 2 = 9x

2=8x2 = 8x

x=14x = \dfrac{1}{4}

Check the domain — both xx and x+2x+2 must be positive. x=14x = \dfrac{1}{4} satisfies this.

Final answer: x=14x = \dfrac{1}{4}.

Where The Log Rules Come From

Every log rule is the same statement read in two languages. In exponent language, bx⋅by=bx+yb^x \cdot b^y = b^{x+y}. In log language, that exact rule reads log⁡b(MN)=log⁡bM+log⁡bN\log_b(MN) = \log_b M + \log_b N. The translation works because the log function is the inverse of the exponential function — whatever exponents do, logs undo, in mirror image.

Deriving the product rule. Let log⁡bM=x\log_b M = x and log⁡bN=y\log_b N = y.

By the definition of a logarithm: M=bxM = b^x and N=byN = b^y.

Multiply:

MN=bx⋅by=bx+yMN = b^x \cdot b^y = b^{x+y}

Take log⁡b\log_b of both sides:

log⁡b(MN)=x+y\log_b(MN) = x + y

Substitute xx and yy back:

log⁡b(MN)=log⁡bM+log⁡bN\log_b(MN) = \log_b M + \log_b N

That is the product rule. The quotient rule follows the same proof with division replacing multiplication and bx−yb^{x-y} replacing bx+yb^{x+y}. The power rule follows from (bx)k=bkx(b^x)^k = b^{kx}.

Deriving change of base. Let log⁡bx=c\log_b x = c, so bc=xb^c = x. Take log⁡a\log_a of both sides:

log⁡a(bc)=log⁡ax\log_a(b^c) = \log_a x

Apply the power rule on the left:

c⋅log⁡ab=log⁡axc \cdot \log_a b = \log_a x

Solve for cc:

c=log⁡axlog⁡abc = \dfrac{\log_a x}{\log_a b}

Substitute back: log⁡bx=log⁡axlog⁡ab\log_b x = \dfrac{\log_a x}{\log_a b}. The reciprocal rule is the special case where a=xa = x.

Why Logarithms Exist at All

Logarithms were invented to turn multiplication into addition. That sounds small. It changed the speed of human science.

Before electronic calculators, every astronomer, navigator, and engineer who needed to multiply large numbers did it by hand. A single multiplication of two seven-digit numbers could take twenty minutes and was easy to get wrong. The astronomer Pierre-Simon Laplace later said that logarithms "by shortening the labours, doubled the life of the astronomer."

The shortcut works for one reason. By the product rule, log⁡(MN)=log⁡M+log⁡N\log(MN) = \log M + \log N. So instead of multiplying two large numbers directly, you could:

  • Look up the log of each number in a printed table.

  • Add the two logs (a quick paper-and-pencil step).

  • Look up the answer's log in reverse — what is called the "antilog."

A multiplication that took twenty minutes became an addition that took thirty seconds. From the early 1600s until the 1970s, every long-distance ship, every census calculation, and every astronomical orbit was computed this way.

Logarithms also matter today, even though calculators do the arithmetic for us. Three reasons:

  • They linearise growth. Quantities that grow exponentially — populations, bank balances, computing power — become straight lines on a log scale. That is why finance and biology graphs are often plotted on logarithmic axes.

  • They measure perception. Human ears, eyes, and chemical sensors respond to ratios of stimulus, not differences. The decibel scale (sound), the Richter scale (earthquakes), the pH scale (acidity), and stellar magnitudes (brightness) are all logarithmic for this reason.

  • They run information theory. Every digital file's compression, every search engine's ranking, every modern cryptographic protocol uses log⁡2\log_2 as the unit of information.

The log rules from this article are what make all of this possible. Every one of them is a property of exponents, written in mirror.

The Mathematicians Behind Logarithms

Three names sit at the start of the log story.

John Napier (1550–1617, Scotland) was a Scottish landowner and theologian who treated mathematics as a private hobby. He spent twenty years - from his late thirties to his late fifties - building the first table of logarithms by hand, working alone in his castle near Edinburgh.

His 1614 book Mirifici Logarithmorum Canonis Descriptio introduced the word logarithm to the world. Napier never lived to see how completely his invention would reshape science. He died three years after publishing the book.

Henry Briggs (1561–1630, England) was the first professor of geometry at Gresham College in London. After reading Napier's book, Briggs travelled to Scotland to meet him. The two reportedly sat in silence for fifteen minutes when they first met, each unable to speak from admiration of the other.

Briggs reworked Napier's tables to base 10 - the common logarithm every student now uses - and published the first base-10 log tables in 1617.

Leonhard Euler (1707–1783, Switzerland) was the eighteenth-century mathematician who connected logarithms to the constant ee and to the exponential function as we now write it. The notation ln⁡x\ln x for the natural logarithm and the modern definition eln⁡x=xe^{\ln x} = x both come from Euler.

[MATHEMATICIANS & HISTORY CALLOUT]

Title: Twenty years of arithmetic, alone

Mathematician: John Napier

Date and place: 1594–1614, Merchiston Castle, near Edinburgh, Scotland

The story: Napier did not know if his idea would work when he started. He spent two decades computing logarithms one entry at a time, with no calculator, no collaborator, and no published precedent. There were rumours during his lifetime that he kept a black rooster as a familiar and that the locals thought he was a wizard. When his table was finally finished and printed in 1614, every astronomer in Europe wanted a copy within months.

Why it matters: A single human, working in obscurity, can build a tool that the rest of the world ends up using for three hundred years. Logarithms were that tool.

Common Mistakes With Log Formulas

1. Splitting a log of a sum. log⁡b(M+N)≠log⁡bM+log⁡bN\log_b(M + N) \neq \log_b M + \log_b N.

Where it slips in: The product rule looks like distribution, so students apply it to addition by reflex. The product rule splits a log of a product, not a log of a sum. There is no clean rule for log⁡(M+N)\log(M+N).

Don't do this: log⁡(3+4)=log⁡3+log⁡4\log(3 + 4) = \log 3 + \log 4.

The correct way: log⁡(3+4)=log⁡7\log(3 + 4) = \log 7 — evaluate the sum first, then take the log if needed.

2. Pulling a coefficient out of the wrong spot.

Where it slips in: The power rule says k⋅log⁡bM=log⁡b(Mk)k \cdot \log_b M = \log_b(M^k) — the coefficient becomes an exponent on the argument MM, not on the whole log. Students sometimes write 2log⁡5=(log⁡5)22 \log 5 = (\log 5)^2 by mistake.

Don't do this: 2log⁡5=(log⁡5)22 \log 5 = (\log 5)^2.

The correct way: 2log⁡5=log⁡52=log⁡252 \log 5 = \log 5^2 = \log 25.

3. Forgetting that the argument must be positive.

Where it slips in: After solving a log equation, students substitute their answer back without checking whether the argument is still positive. An algebraically valid answer can be a domain failure.

Don't do this: Accept x=−3x = -3 as a solution to log⁡(x+1)=log⁡(−2x−5)\log(x + 1) = \log(-2x - 5) without checking.

The correct way: For every solution, verify each log⁡b(argument)\log_b(\text{argument}) has its argument >0> 0. Reject any solution that breaks this.

4. Confusing log⁡b(MN)\log_b(MN) with (log⁡bM)(log⁡bN)(\log_b M)(\log_b N).

Where it slips in: The product rule turns a log of a product into a sum of logs, not a product of logs. Multiplication of two logs has no simple rule.

Don't do this: log⁡2(3⋅5)=(log⁡23)(log⁡25)\log_2(3 \cdot 5) = (\log_2 3)(\log_2 5).

The correct way: log⁡2(3⋅5)=log⁡23+log⁡25\log_2(3 \cdot 5) = \log_2 3 + \log_2 5.

A real-world version of the wrong-base mistake

Chemistry students computing pH — defined as pH=−log⁡10[H+]\text{pH} = -\log_{10}[\text{H}^+] — sometimes use ln⁡\ln instead of log⁡\log on the calculator. The result is off by a factor of about 2.3, because ln⁡10≈2.303\ln 10 \approx 2.303. A 2.3-unit pH error is the difference between drinking water (pH 7) and battery acid (pH 0–1). The same single-character base mismatch — log⁡\log versus ln⁡\ln — has shown up in pharmaceutical dosing calculations, in audio engineering decibel conversions, and in seismograph readings. The rule that prevents it is the same one as in the third common mistake above: check the base every time.

Formula

Where it comes from

log⁡b!(1N)=−log⁡bN\log_b!\left(\dfrac{1}{N}\right) = -\log_b N

Quotient rule with M=1M = 1

log⁡bMn=1nlog⁡bM\log_b \sqrt[n]{M} = \dfrac{1}{n}\log_b M

Power rule with k=1nk = \dfrac{1}{n}

ln⁡x=log⁡ex\ln x = \log_e x

Natural log — base ee special case

log⁡x=log⁡10x\log x = \log_{10} x

Common log — base 10 special case

log⁡b(bk)=k\log_b(b^k) = k

Inverse rule — logs and exponentials cancel

log⁡ba⋅log⁡ab=1\log_b a \cdot \log_a b = 1

Reciprocal rule, rewritten

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Frequently Asked Questions

Where do log formulas come from?
From the rules of exponents. Each log formula is the same statement as one of the exponent rules, rewritten with the base and exponent swapped. The product rule of logs comes from bx⋅by=bx+yb^x \cdot b^y = b^{x+y}. The power rule of logs comes from (bx)k=bkx(b^x)^k = b^{kx}. Memorising the exponent rules and the meaning of log⁡ba=c  ⟺  bc=a\log_b a = c \iff b^c = a is enough to derive every log rule from scratch.
What is the difference between log⁡\log and ln⁡\ln?
log⁡x\log x usually means log⁡10x\log_{10} x — the common logarithm, base 10. ln⁡x\ln x means log⁡ex\log_e x — the natural logarithm, base ee where e≈2.71828e \approx 2.71828. They differ only in base. Every rule on this page works for either. Calculators have separate buttons for both because both come up constantly in engineering, science, and finance.
Can the argument of a log be negative or zero?
No, not in real numbers. log⁡b(0)\log_b(0) is undefined because no power of bb equals zero. log⁡b\log_b of a negative number is also undefined in real numbers because bcb^c is always positive when b>0b > 0. (In complex numbers the situation is different, but that is a topic for later coursework.)
Why is log⁡b1=0\log_b 1 = 0 for every base?
Because b0=1b^0 = 1 for every nonzero base. Logarithms ask *what power of bb gives the argument?* The argument is 1, the answer is 0, regardless of bb.
Are log formulas in the school syllabus?
Yes. Logarithms typically appear in Grade 10 or 11 algebra in most curricula — for instance, around CCSS HSF.LE.A.4 in the United States and Class 11 mathematics in the Indian NCERT curriculum. They reappear in calculus, in chemistry (pH and reaction kinetics), in physics (decibels, magnitudes), and in finance (compound interest and continuous growth).
✍️ Written By
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Bhanzu Team
Content Creator and Editor
Bhanzu’s editorial team, known as Team Bhanzu, is made up of experienced educators, curriculum experts, content strategists, and fact-checkers dedicated to making math simple and engaging for learners worldwide. Every article and resource is carefully researched, thoughtfully structured, and rigorously reviewed to ensure accuracy, clarity, and real-world relevance. We understand that building strong math foundations can raise questions for students and parents alike. That’s why Team Bhanzu focuses on delivering practical insights, concept-driven explanations, and trustworthy guidance-empowering learners to develop confidence, speed, and a lifelong love for mathematics.
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