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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

by=xb^y=x

with

b>0b>0

and

b1b\neq1

, then

logbx=y\log_b x = y

. That single identity—log ↔ exponent—is the thread that ties every rule together. When you read

log101000=3\log_{10}1000=3

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

ln(e2.5)=2.5\ln(e^{2.5})=2.5

, you’re seeing the natural log “undo” the exponential

e2.5e^{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>0x>0

). Bases must satisfy

b>0b>0

and

b1b\neq1

. 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

by=x    logbx=yb^y=x \iff \log_b x = y

, we obtain the following core rules for

x>0, y>0, b>0, b1x>0,\ y>0,\ b>0,\ b\neq1

:

  • Product rule:

    logb(xy)=logbx+logby\log_b(xy)=\log_b x+\log_b y
    (exponent on a product is a sum of exponents)

  • Quotient rule:

    logb ⁣(xy)=logbxlogby\log_b\!\left(\frac{x}{y}\right)=\log_b x-\log_b y

  • Power rule:

    logb(xr)=rlogbx\log_b(x^r)=r\log_b x for any real

    rr

  • Root as power:

    logb ⁣(xr)=1rlogbx\log_b\!\left(\sqrt[r]{x}\right)=\frac{1}{r}\log_b x

  • Inverse rules:

    blogbx=xb^{\log_b x}=x and

    logb(by)=y\log_b(b^y)=y

  • Base conversions (preview):

    logbx=logkxlogkb\log_b x=\dfrac{\log_k x}{\log_k b} for any valid base

    kk (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  

logb(xy)=logbx+logby\log_b(xy)=\log_b x+\log_b y

Turn multiplication inside into addition outside
Quotient  

logb ⁣(xy)=logbxlogby\log_b\!\left(\frac{x}{y}\right)=\log_b x-\log_b y

Turn division inside into subtraction outside
Power  

logb(xr)=rlogbx\log_b(x^r)=r\log_b x

Pull powers down to simplify or solve
Root  

logb(xr)=1rlogbx\log_b(\sqrt[r]{x})=\frac{1}{r}\log_b x

Roots become fractions of logs
Inverses  

blogbx=x, logb(by)=yb^{\log_b x}=x,\ \log_b(b^y)=y

Log and exponential undo each other
Change of base  

logbx=logkxlogkb\log_b x=\dfrac{\log_k x}{\log_k b}

Evaluate any log using a convenient base

Worked micro-examples (properties in action):

  • Product:

    log2(84)=log28+log24=3+2=5.\log_2(8\cdot4)=\log_2 8+\log_2 4=3+2=5.

  • Quotient:

    log3 ⁣(813)=log381log33=41=3.\log_3\!\left(\frac{81}{3}\right)=\log_3 81-\log_3 3=4-1=3.

  • Power:

    log5(253)=3log525=32=6.\log_5(25^3)=3\log_5 25=3\cdot2=6.

  • Inverse:

    10log107=710^{\log_{10}7}=7 and

    log10(101.2)=1.2.\log_{10}(10^{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

logbx\log_b x

on any calculator—even when it doesn’t have a dedicated base

bb

key. For any valid base

kk

,

 

logbx=logkxlogkb.\log_b x=\frac{\log_k x}{\log_k b}.

Two popular choices are

k=10k=10

(common log) and

k=ek=e

(natural log). Both give the same value:

 

logbx=log10xlog10b=lnxlnb.\log_b x=\frac{\log_{10}x}{\log_{10}b}=\frac{\ln x}{\ln b}.

Derivation (short and memorable): Let

y=logbxy=\log_b x

. By definition,

by=xb^y=x

. Apply

logk\log_k

to both sides:

 

logk(by)=logkxylogkb=logkxy=logkxlogkb.\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}.

Because

y=logbxy=\log_b x

, we’re done.

Calculator workflow: To evaluate

log750\log_7 50

, enter

ln50÷ln7\ln 50 ÷ \ln 7

. On many devices, the order

(ln50)/(ln7)(\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,

log7503.9120230051.945910149=2.009\log_7 50\approx\frac{3.912023005}{1.945910149}=2.009…

. Rounding only at the end protects your final answer.

Strategic uses beyond evaluation:

  • Comparing magnitudes: If you need to compare

    log3(n)\log_{3}(n) and

    log5(n)\log_{5}(n) for large

    nn, change to a common base and note that

    log3n=lnnln3\log_3 n=\dfrac{\ln n}{\ln 3} and

    log5n=lnnln5\log_5 n=\dfrac{\ln n}{\ln 5}. Since

    ln3<ln5\ln 3<\ln 5,

    log3n>log5n\log_{3} n>\log_{5} n for all

    n>0n>0.

  • Simplifying expressions with mixed bases: Expressions like

    log2(x)+log5(x)\log_{2}(x)+\log_{5}(x) become

    (1ln2+1ln5)lnx\left(\frac{1}{\ln2}+\frac{1}{\ln5}\right)\ln x after changing to base

    ee. 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

  1. Combine logs first, then exponentiate.
    Solve

    log3(x)+log3(x2)=2\log_3(x)+\log_3(x-2)=2. Product rule gives

    log3(x(x2))=2\log_3\big(x(x-2)\big)=2. Exponentiate base 3:

    x(x2)=32=9x22x9=0x(x-2)=3^2=9\Rightarrow x^2-2x-9=0. The roots are

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

    x>2x>2, so

    x=1+104.162x=1+\sqrt{10}\approx4.162 is valid;

    x=110<0x=1-\sqrt{10}<0 is rejected.

  2. Use the power rule to pull down exponents.
    Solve

    2log10(x)=32\log_{10}(x)=3. Power rule gives

    log10(x2)=3\log_{10}(x^2)=3. Then

    x2=103=1000x=100031.622x^2=10^3=1000\Rightarrow x=\sqrt{1000}\approx31.622. The domain

    x>0x>0 means

    x31.622x\approx31.622 only.

  3. Equations with different log bases.
    Solve

    log2(x)=log5(20)\log_{2}(x)=\log_{5}(20). Apply change-of-base to each side using natural logs:

    lnxln2=ln20ln5lnx=ln2ln20ln5\frac{\ln x}{\ln 2}=\frac{\ln 20}{\ln 5}\Rightarrow \ln x=\ln 2\cdot \frac{\ln 20}{\ln 5}. Exponentiate to get

 

x=exp ⁣(ln2ln20ln5)=2log520.x=\exp\!\left(\ln 2\cdot \frac{\ln 20}{\ln 5}\right)=2^{\log_{5}20}.

You can evaluate numerically (

x21.8613.62x\approx 2^{1.861…}\approx3.62

) or keep the exact exponent form.

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

  1. Isolate the exponential and log both sides.
    Solve

    52x1=405^{2x-1}=40. Take natural log:

    (2x1)ln5=ln402x1=ln40ln5x=12 ⁣(ln40ln5+1)12(1.861+1)=1.4305.(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. Different bases? No problem.
    Solve

    3x=73^x=7. Then

    x=ln7ln31.7712x=\dfrac{\ln7}{\ln3}\approx1.7712. This is the quintessential change-of-base application.

  3. Exponential expressions hiding as products.
    If

    axbx=ca^{x}\cdot b^{x}=c, rewrite as

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

    ab>0ab>0 and

    ab1ab\neq1. Small algebraic rewrites can save time and reduce errors.

C. Modeling quick hits: In growth/decay

P(t)=P0ektP(t)=P_0e^{kt}

, the time to reach a level

LL

is

t=1kln ⁣(LP0)t=\dfrac{1}{k}\ln\!\left(\frac{L}{P_0}\right)

. 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

logbx\log_b x

, the argument must be positive. When solving

log2(x5)=3\log_2(x-5)=3

, the solution

x=13x=13

is fine, but if intermediate algebra offered

x=4x=4

you must reject it because

x5x-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

logb(x+y)\log_b(x+y)

into

logbx+logby\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

logb(xr)=rlogbx\log_b(x^r)=r\log_b x

. The reverse is

logbx=1rlogb(xr)\log_b x = \frac{1}{r}\log_b(x^r)

, not

logb(rx)\log_b(rx)

. Multiplying the argument by

rr

is not the same as raising it to a power.

4) Forgetting base constraints.
Log bases must satisfy

b>0,b1b>0, b\neq1

. Writing

log2x\log_{-2}x

or

log1x\log_1 x

is undefined. When changing base, choose any valid

kk

(usually

ee

or

1010

) 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,

log750\log_7 50

stabilizes if you store full precision for

ln50\ln 50

and

ln7\ln 7

before dividing.

6) Ignoring the inverse relationship.
The pair

logb\log_b

and

bxb^{x}

are inverses. If an equation contains both, consider applying one to undo the other. For instance, if you have

blogbf(x)b^{\log_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

log10\log_{10}

.

lnx\ln x

uses base

e2.71828e\approx2.71828

and

logx\log x

often means base

1010

in many contexts. In higher math,

log\log

sometimes defaults to base

ee

. State your base or stick to explicit notation like

log10\log_{10}

and

ln\ln

to avoid miscommunication.

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

E=log3 ⁣(2793)E=\log_3\!\left(\dfrac{27\sqrt{9}}{3}\right)

. Inside the log,

27=3327=3^3

,

9=3\sqrt{9}=3

, and the denominator is

33

. So

E=log3 ⁣(3333)=log3(33)=3E=\log_3\!\left(\dfrac{3^3\cdot3}{3}\right)=\log_3(3^3)=3

. Notice how product/quotient rules plus recognizing perfect powers make the calculation nearly mental.

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