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Mathematics AI · SL · Unit 1: Number and Algebra

Unit 1Topic 1.5Key Concepts51 min read

Integer exponents and logarithms

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Integer exponent foundations

An exponent tells us how a base is used as a factor. For a positive integer nn, the expression ana^n means a product of nn factors, each equal to aa. For example, 34=3×3×3×3=813^4=3\times3\times3\times3=81. The base is 33, the exponent is 44, and the power is the complete expression 343^4.

The exponent applies only to the quantity immediately before it unless brackets extend its scope. Thus −24=−(24)=−16-2^4=-(2^4)=-16, but (−2)4=16(-2)^4=16. With a negative base, an even integer exponent gives a positive result and an odd integer exponent gives a negative result.

For every non-zero base, a0=1a^0=1. This follows from the quotient law: amam=am−m=a0\frac{a^m}{a^m}=a^{m-m}=a^0, while the fraction itself equals 11. The restriction a≠0a\ne0 matters: 000^0 is not assigned the ordinary value 11 in this course.

A negative exponent creates a reciprocal: a−n=1ana^{-n}=\frac{1}{a^n} for a≠0a\ne0. The minus sign belongs to the exponent; it does not make the whole value negative. For instance, 2−3=123=182^{-3}=\frac{1}{2^3}=\frac{1}{8}, which is positive.

Key terms

Base
The repeated factor in a power. In ana^n, the base is aa.
Exponent
The number that describes how the base is powered. In ana^n, the exponent is nn.
Power
The complete expression consisting of a base and exponent, such as 535^3.
Reciprocal
The multiplicative inverse of a non-zero number. The reciprocal of aa is 1a\frac{1}{a}.
Integer exponent
An exponent in Z\mathbb Z: a positive whole number, zero, or a negative whole number.

Worked examples

Interpret positive, zero, and negative exponents

Evaluate 434^3, 707^0, and 5−25^{-2}.

  1. 1Use repeated multiplication: 43=4×4×4=644^3=4\times4\times4=64.
  2. 2Because 7≠07\ne0, use the zero-exponent rule: 70=17^0=1.
  3. 3Use the reciprocal rule: 5−2=152=1255^{-2}=\frac{1}{5^2}=\frac{1}{25}.

Answer: 43=644^3=64, 70=17^0=1, and 5−2=1255^{-2}=\frac{1}{25}.

Read brackets before evaluating

Compare −32-3^2 and (−3)2(-3)^2.

  1. 1In −32-3^2, the exponent applies to 33 only, so −32=−(32)-3^2=-(3^2).
  2. 2Therefore −32=−9-3^2=-9.
  3. 3In (−3)2(-3)^2, brackets make −3-3 the base.
  4. 4Therefore (−3)2=(−3)(−3)=9(-3)^2=(-3)(-3)=9.

Answer: −32=−9-3^2=-9, whereas (−3)2=9(-3)^2=9.

Rules and formulas

Positive integer exponent

an=a⋅a⋯a⏟n factors,n∈Na^n=\underbrace{a\cdot a\cdots a}_{n\text{ factors}},\quad n\in\mathbb N
aa
the base
nn
a positive integer exponent

Zero exponent

a0=1,a≠0a^0=1,\quad a\ne0

Negative integer exponent

a−n=1an,a≠0a^{-n}=\frac{1}{a^n},\quad a\ne0

Laws of integer exponents

The exponent laws are consequences of counting equal factors. When powers with the same non-zero base are multiplied, add their exponents. When they are divided, subtract the denominator exponent from the numerator exponent. These rules work for positive, zero, and negative integer exponents whenever the expressions are defined.

A power raised to another power multiplies the exponents. A product or quotient raised to a power applies that power to every factor in the brackets. The same-base condition is essential for the product and quotient laws: 23×532^3\times5^3 cannot be simplified by adding the exponents, although it can be written as (2×5)3=103(2\times5)^3=10^3.

Exponent laws change products, quotients, and powers—not sums. In general, (a+b)n≠an+bn(a+b)^n\ne a^n+b^n. For example, (2+3)2=25(2+3)^2=25, while 22+32=132^2+3^2=13.

Key terms

Product law
For the same base, multiplication adds exponents: aman=am+na^m a^n=a^{m+n}.
Quotient law
For the same non-zero base, division subtracts exponents: aman=am−n\frac{a^m}{a^n}=a^{m-n}.
Power of a power
Multiply the exponents: (am)n=amn(a^m)^n=a^{mn}.

Worked examples

Multiply and divide powers

Simplify x7x−3x2\frac{x^7x^{-3}}{x^2}, where x≠0x\ne0.

  1. 1Multiplication adds the first two exponents: x7x−3=x7+(−3)=x4x^7x^{-3}=x^{7+(-3)}=x^4.
  2. 2Division subtracts the denominator exponent: x4x2=x4−2\frac{x^4}{x^2}=x^{4-2}.

Answer: x2x^2.

Apply a power to a quotient

Simplify (2a3b−2)2\left(\frac{2a^3}{b^{-2}}\right)^2, where a,b≠0a,b\ne0.

  1. 1First note that dividing by b−2b^{-2} is multiplying by b2b^2, so the bracket is 2a3b22a^3b^2.
  2. 2Square every factor: 22(a3)2(b2)22^2(a^3)^2(b^2)^2.
  3. 3Use the power-of-a-power law: (a3)2=a6(a^3)^2=a^6 and (b2)2=b4(b^2)^2=b^4.

Answer: 4a6b44a^6b^4.

Rules and formulas

Product law

aman=am+na^m a^n=a^{m+n}

Quotient law

aman=am−n,a≠0\frac{a^m}{a^n}=a^{m-n},\quad a\ne0

Power of a power

(am)n=amn(a^m)^n=a^{mn}

Power of a product

(ab)n=anbn(ab)^n=a^n b^n

Power of a quotient

(ab)n=anbn,b≠0\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n},\quad b\ne0

Simplifying expressions with powers

A strong simplification method is to work factor by factor. Expand only when it helps, collect numerical coefficients, combine each repeated base using exponent laws, and then rewrite negative exponents as reciprocals if the question asks for positive exponents.

An algebraic expression with a negative exponent is not automatically wrong. However, IB questions often request an answer with positive exponents. Move a factor across the fraction bar to change the sign of its exponent: x−4y−2=y2x4\frac{x^{-4}}{y^{-2}}=\frac{y^2}{x^4}.

Restrictions should be carried with algebra involving division or negative exponents. If the original expression contains 1x\frac{1}{x}, then x≠0x\ne0, even if later cancellation removes the visible denominator.

Key terms

Positive-exponent form
An equivalent expression in which no variable has a negative exponent.
Restriction
A condition such as x≠0x\ne0 required to keep an expression defined.
Equivalent expression
A different-looking expression with the same value for every input in the permitted domain.

Worked examples

Simplify a multi-factor expression

Simplify (3x−2y3)29xy−1\frac{(3x^{-2}y^3)^2}{9xy^{-1}} using positive exponents, where x,y≠0x,y\ne0.

  1. 1Square the numerator: (3x−2y3)2=9x−4y6(3x^{-2}y^3)^2=9x^{-4}y^6.
  2. 2Divide by 9xy−19xy^{-1}: the coefficient becomes 99=1\frac{9}{9}=1.
  3. 3Subtract exponents for xx: −4−1=−5-4-1=-5.
  4. 4Subtract exponents for yy: 6−(−1)=76-(-1)=7.
  5. 5Rewrite x−5x^{-5} in the denominator.

Answer: y7x5\frac{y^7}{x^5}.

Simplify before substituting

Given p=2p=2 and q=4q=4, evaluate p−2q3q−1\frac{p^{-2}q^3}{q^{-1}}.

  1. 1Combine the powers of qq: q3q−1=q3−(−1)=q4\frac{q^3}{q^{-1}}=q^{3-(-1)}=q^4.
  2. 2Rewrite p−2=1p2p^{-2}=\frac{1}{p^2}, giving q4p2\frac{q^4}{p^2}.
  3. 3Substitute: 4422=2564\frac{4^4}{2^2}=\frac{256}{4}.

Answer: 6464.

Rules and formulas

Moving a negative power across a fraction bar

a−mb−n=bnam,a,b≠0\frac{a^{-m}}{b^{-n}}=\frac{b^n}{a^m},\quad a,b\ne0

Logarithms as inverse exponents

A logarithm answers one precise question: what exponent must the base be raised to in order to obtain this number? The statement log⁡ab=c\log_a b=c means exactly the same thing as ac=ba^c=b. Logarithms and exponentiation are inverse operations, just as subtraction reverses addition.

Two equivalent statements: two cubed equals eight and logarithm base two of eight equals three.The base, exponent and result do not change when we switch forms: 23=8⟺log⁡2(8)=32^3=8\Longleftrightarrow\log_2(8)=3.

For real logarithms, the base must satisfy a>0a>0 and a≠1a\ne1, and the argument must satisfy b>0b>0. A positive base never produces zero or a negative result when raised to a real power, so log⁡a0\log_a 0 and log⁡a(−5)\log_a(-5) are undefined in the real number system.

Two bases are especially common. The common logarithm log⁡x\log x usually means log⁡10x\log_{10}x. The natural logarithm ln⁡x\ln x means log⁡ex\log_e x, where e≈2.71828e\approx2.71828. Both are calculator functions and both obey the same logarithm laws.

Key terms

Logarithm
The exponent to which a valid base must be raised to produce a positive argument.
Argument
The positive quantity inside a logarithm. In log⁡ab\log_a b, the argument is bb.
Common logarithm
A logarithm with base 1010, usually written log⁡x\log x.
Natural logarithm
A logarithm with base ee, written ln⁡x\ln x.
Inverse operations
Operations that undo one another; exponentiation and logarithms form an inverse pair.

Worked examples

Convert exponential form to logarithmic form

Rewrite 25=322^5=32 in logarithmic form.

  1. 1Identify the base: 22.
  2. 2Identify the exponent: 55.
  3. 3Identify the result or logarithm argument: 3232.
  4. 4Use ac=b⟺log⁡ab=ca^c=b\Longleftrightarrow\log_a b=c.

Answer: log⁡232=5\log_2 32=5.

Convert logarithmic form to exponential form

Rewrite log⁡3(127)=−3\log_3(\frac{1}{27})=-3 in exponential form.

  1. 1The logarithm base is 33.
  2. 2The logarithm value is the exponent −3-3.
  3. 3The argument 127\frac{1}{27} is the result.

Answer: 3−3=1273^{-3}=\frac{1}{27}.

Evaluate exact logarithms

Evaluate log⁡10(0.001)\log_{10}(0.001) and ln⁡(e4)\ln(e^4).

  1. 1Write 0.001=10−30.001=10^{-3}, so the required exponent is −3-3.
  2. 2The exponent that turns ee into e4e^4 is 44.

Answer: log⁡10(0.001)=−3\log_{10}(0.001)=-3 and ln⁡(e4)=4\ln(e^4)=4.

Rules and formulas

Exponential-logarithmic equivalence

ac=b⟺log⁡ab=ca^c=b\quad\Longleftrightarrow\quad\log_a b=c
aa
the logarithm base, with a>0a>0 and a≠1a\ne1
bb
the positive argument
cc
the exponent or logarithm value

Common and natural logarithms

log⁡x=log⁡10x,ln⁡x=log⁡ex\log x=\log_{10}x,\qquad\ln x=\log_e x

Inverse identities

log⁡a(ax)=x,alog⁡ax=x(x>0)\log_a(a^x)=x,\qquad a^{\log_a x}=x\quad(x>0)

Basic logarithm rules and properties

Logarithm laws turn multiplication into addition, division into subtraction, and powers into coefficients. They follow directly from the exponent laws. For example, if x=amx=a^m and y=any=a^n, then xy=am+nxy=a^{m+n}, so log⁡a(xy)=m+n=log⁡ax+log⁡ay\log_a(xy)=m+n=\log_a x+\log_a y.

Every logarithm argument involved in a rule must be positive, and the base must be positive and not equal to 1. There is no general identity log⁡a(x+y)=log⁡ax+log⁡ay\log_a(x+y)=\log_a x+\log_a y. The expressions can coincide for special values, but logarithms do not generally distribute over addition or subtraction.

The power rule is useful in both directions. It can expand log⁡a(xk)\log_a(x^k) to klog⁡axk\log_a x, or condense klog⁡axk\log_a x to log⁡a(xk)\log_a(x^k), provided the domain conditions are satisfied.

Key terms

Product property
For x,y>0x,y>0 and a>0, a≠1a>0,\ a\ne1, log⁡a(xy)=log⁡ax+log⁡ay\log_a(xy)=\log_a x+\log_a y. Each factor, not just the product, must be positive.
Quotient property
For x,y>0x,y>0 and a>0, a≠1a>0,\ a\ne1, log⁡a(xy)=log⁡ax−log⁡ay\log_a(\frac{x}{y})=\log_a x-\log_a y. Both numerator and denominator must be positive.
Power property
An exponent on a positive argument can be written as a coefficient of the logarithm.
Condense
Combine a sum or difference of logarithms into one logarithm.
Expand
Rewrite one logarithm of a product, quotient, or power as several simpler terms.

Worked examples

Expand a logarithmic expression

Expand log⁡2(8x3y)\log_2(\frac{8x^3}{y}), where x,y>0x,y>0.

  1. 1Use the quotient law: log⁡2(8x3)−log⁡2y\log_2(8x^3)-\log_2 y.
  2. 2Use the product law: log⁡28+log⁡2(x3)−log⁡2y\log_2 8+\log_2(x^3)-\log_2 y.
  3. 3Use the power law: log⁡28+3log⁡2x−log⁡2y\log_2 8+3\log_2x-\log_2y.
  4. 4Evaluate log⁡28=3\log_2 8=3.

Answer: 3+3log⁡2x−log⁡2y3+3\log_2x-\log_2y.

Condense logarithms

Write 2ln⁡x−ln⁡y2\ln x-\ln y as one logarithm, where x,y>0x,y>0.

  1. 1Use the power law backwards: 2ln⁡x=ln⁡(x2)2\ln x=\ln(x^2).
  2. 2Use the quotient law backwards: ln⁡(x2)−ln⁡y=ln⁡(x2y)\ln(x^2)-\ln y=\ln(\frac{x^2}{y}).

Answer: ln⁡(x2y)\ln\left(\frac{x^2}{y}\right).

Rules and formulas

Product property

log⁡a(xy)=log⁡ax+log⁡ay\log_a(xy)=\log_a x+\log_a y

Quotient property

log⁡a(xy)=log⁡ax−log⁡ay\log_a\left(\frac{x}{y}\right)=\log_a x-\log_a y

Power property

log⁡a(xk)=klog⁡ax\log_a(x^k)=k\log_a x

Useful exact values

log⁡a1=0,log⁡aa=1\log_a1=0,\qquad\log_a a=1

Solving exponential equations

First look for a common base. If both sides can be written as powers of the same valid base, equate the exponents. This gives an exact solution and is usually the fastest non-calculator method.

If a common base is not convenient, isolate the exponential expression and take a logarithm of both sides. Either ln⁡\ln or log⁡\log works because the change-of-base ratio cancels consistently. Keep full calculator precision until the final answer.

In contextual problems, the algebraic value may need interpretation. If time is counted in complete years or cycles and a quantity must first exceed a target, test the neighbouring integers rather than simply rounding the logarithmic answer.

Key terms

Common-base method
Rewrite both sides with the same base and equate the exponents.
Logarithmic method
Take logarithms after isolating an exponential term, then use the power property to bring down the variable exponent.
Exact solution
A value written without rounding, such as x=3x=3 or x=ln⁡7ln⁡2x=\frac{\ln7}{\ln2}.

Worked examples

Solve by matching bases

Solve 52x−1=1255^{2x-1}=125.

  1. 1Write 125=53125=5^3.
  2. 2The bases now match: 52x−1=535^{2x-1}=5^3.
  3. 3Equate exponents: 2x−1=32x-1=3.
  4. 4Solve the linear equation: 2x=42x=4.

Answer: x=2x=2.

Solve using logarithms

Solve 2x=152^x=15, giving the answer to 3 significant figures.

  1. 1Take natural logarithms: ln⁡(2x)=ln⁡15\ln(2^x)=\ln15.
  2. 2Use the power property: xln⁡2=ln⁡15x\ln2=\ln15.
  3. 3Divide: x=ln⁡15ln⁡2x=\frac{\ln15}{\ln2}.
  4. 4Evaluate with full calculator precision, then round.

Answer: x≈3.91x\approx3.91 (3 s.f.).

Interpret a growth threshold

A population is 120(1.18)t120(1.18)^t. Find the first whole period for which it exceeds 300300.

  1. 1Solve the boundary equation 120(1.18)t=300120(1.18)^t=300.
  2. 2Divide by 120120: (1.18)t=2.5(1.18)^t=2.5.
  3. 3Take logarithms: t=ln⁡2.5ln⁡1.18≈5.54t=\frac{\ln2.5}{\ln1.18}\approx5.54.
  4. 4The first complete period after 5.545.54 is 66.
  5. 5Check that period 5 is below and period 6 is above the target.

Answer: The population first exceeds 300300 after 66 complete periods.

Rules and formulas

Common-base principle

af(x)=ag(x)⟹f(x)=g(x),a>0, a≠1a^{f(x)}=a^{g(x)}\Longrightarrow f(x)=g(x),\quad a>0,\ a\ne1

Solving a basic exponential equation

ax=b⟹x=ln⁡bln⁡a,a>0, a≠1, b>0a^x=b\Longrightarrow x=\frac{\ln b}{\ln a},\quad a>0,\ a\ne1,\ b>0

Solving simple logarithmic equations

For a single logarithm, convert directly to exponential form. For several logarithms with the same base, first condense them using the logarithm laws, then convert to exponential form. Solve the resulting algebraic equation and check every candidate in the original logarithms.

Domain checks are part of the solution. A candidate that makes any logarithm argument zero or negative must be rejected, even if it satisfies the transformed polynomial equation.

When logarithms appear on both sides with the same one-to-one base, equal logarithms of valid positive arguments imply equal arguments: log⁡af(x)=log⁡ag(x)⇒f(x)=g(x)\log_a f(x)=\log_a g(x)\Rightarrow f(x)=g(x), subject to both domain conditions.

Key terms

Extraneous solution
A value produced by algebraic manipulation that is invalid in the original equation, often because it violates a logarithm domain.
Domain check
Substitution into every original logarithm argument to confirm it is strictly positive.
One-to-one
A property ensuring equal outputs of a logarithmic function come from equal valid inputs.

Worked examples

Solve one logarithm

Solve log⁡3(x−1)=2\log_3(x-1)=2.

  1. 1Require x−1>0x-1>0, so x>1x>1.
  2. 2Convert to exponential form: x−1=32x-1=3^2.
  3. 3Solve: x−1=9x-1=9, so x=10x=10.
  4. 4Check: x−1=9>0x-1=9>0 and log⁡39=2\log_3 9=2.

Answer: x=10x=10.

Condense and reject an invalid root

Solve log⁡x+log⁡(x−9)=1\log x+\log(x-9)=1, where log⁡\log has base 10.

  1. 1Require x>0x>0 and x−9>0x-9>0, so x>9x>9.
  2. 2Use the product law: log⁡[x(x−9)]=1\log[x(x-9)]=1.
  3. 3Convert to exponential form: x(x−9)=101=10x(x-9)=10^1=10.
  4. 4Solve x2−9x−10=0x^2-9x-10=0: (x−10)(x+1)=0(x-10)(x+1)=0.
  5. 5The candidates are 1010 and −1-1. Reject −1-1 because it violates x>9x>9.

Answer: x=10x=10 only.

Use equal logarithms

Solve ln⁡(2x+3)=ln⁡(5x−6)\ln(2x+3)=\ln(5x-6).

  1. 1Require 2x+3>02x+3>0 and 5x−6>05x-6>0, so in particular x>65x>\frac{6}{5}.
  2. 2Because ln⁡\ln is one-to-one, set 2x+3=5x−62x+3=5x-6.
  3. 3Solve: 9=3x9=3x, giving x=3x=3.
  4. 4Both original arguments are positive at x=3x=3.

Answer: x=3x=3.

Rules and formulas

Single logarithm equation

log⁡af(x)=c⟺f(x)=ac,f(x)>0\log_a f(x)=c\Longleftrightarrow f(x)=a^c,\quad f(x)>0

Equality of logarithms

log⁡af(x)=log⁡ag(x)⟹f(x)=g(x),f(x),g(x)>0\log_a f(x)=\log_a g(x)\Longrightarrow f(x)=g(x),\quad f(x),g(x)>0

Exponential growth and decay applications

An exponential model has the form Q(t)=Q0btQ(t)=Q_0b^t, where Q0Q_0 is the initial value and bb is the multiplier per time period. Growth has b>1b>1; decay has 0<b<10<b<1. A rate of r%r\% gives b=1+r100b=1+\frac{r}{100} for growth and b=1−r100b=1-\frac{r}{100} for decay.

Always match the time variable to the rate period. If a population grows by 8%8\% each hour, then tt must count hours in Q=Q0(1.08)tQ=Q_0(1.08)^t. If time is given in minutes, convert it to hours before substituting unless the model defines a minute-by-minute multiplier.

Logarithms answer inverse questions in a model. Substitution finds the quantity at a known time; solving Q0bt=TQ_0b^t=T finds the time required to reach a target: t=ln⁡(TQ0)ln⁡bt=\frac{\ln(\frac{T}{Q_0})}{\ln b}, provided all quantities are positive and b≠1b\ne1.

A mathematical answer still needs contextual interpretation. A time of 4.414.41 years may mean approximately 4.414.41 years for a continuous model, but if a value is checked only once per whole year, test years 44 and 55 and report the first year that satisfies the condition. Exponential models also assume a constant multiplier; this may become unrealistic over long intervals.

Key terms

Initial value
The value Q0Q_0 when t=0t=0.
Growth factor
The multiplier 1+r1+r when the decimal growth rate is r>0r>0.
Decay factor
The multiplier 1−r1-r when the decimal decay rate is rr and 0<r<10<r<1.
Threshold
A target level that a modeled quantity must reach, exceed or fall below.

Worked examples

Predict exponential growth

A bacteria culture starts with 650650 cells and grows by 8%8\% each hour. Estimate the population after 1212 hours.

  1. 1The growth factor is 1+0.08=1.081+0.08=1.08.
  2. 2Use N=650(1.08)12N=650(1.08)^{12}.
  3. 3The GDC gives N≈1636.8106N\approx1636.8106. A cell count must be a whole number.

Answer: Approximately 16371637 cells.

Predict exponential decay

A medicine dose is modeled by A=80(0.72)tA=80(0.72)^t milligrams, where tt is measured in hours. Find the amount after 66 hours.

  1. 1Substitute t=6t=6: A=80(0.72)6A=80(0.72)^6.
  2. 2Evaluate with full precision: A≈11.14512556A\approx11.14512556.
  3. 3Round only the final value and include the unit.

Answer: A≈11.1 mgA\approx11.1\text{ mg} to 3 significant figures.

Find when a value falls below a target

A machine worth $24 000\$24\,000 depreciates according to V=24000(0.82)tV=24000(0.82)^t. Find when its value first falls below $10 000\$10\,000.

  1. 1Solve the boundary equation 24000(0.82)t=1000024000(0.82)^t=10000.
  2. 2Divide by 2400024000: (0.82)t=1000024000(0.82)^t=\frac{10000}{24000}.
  3. 3Take logarithms: t=ln⁡(1000024000)ln⁡(0.82)≈4.4115t=\frac{\ln(\frac{10000}{24000})}{\ln(0.82)}\approx4.4115.
  4. 4Because 0.82<10.82<1, the value is below the target for t>4.4115t>4.4115. If values are recorded at the end of each whole year, test years 44 and 55.

Answer: After about 4.414.41 years, or first at the end of year 55 if only whole years are allowed.

Rules and formulas

Exponential growth or decay model

Q(t)=Q0btQ(t)=Q_0b^t
Q0Q_0
initial value
bb
multiplier per period
tt
number of matching time periods

Percentage multiplier

b=1±r100b=1\pm\frac{r}{100}
rr
percentage rate per period

Time to a target

t=ln⁡(TQ0)ln⁡bt=\frac{\ln(\frac{T}{Q_0})}{\ln b}
TT
positive target value

Logarithmic scales in context

Logarithmic scales compress quantities that vary across many powers of 1010. A change of 11 unit on a base-10 logarithmic scale represents multiplication by 1010 in the original quantity. This is why logarithms are useful for acidity, sound intensity and earthquake measurements.

The pH model is pH=−log⁡10[H+]\mathrm{pH}=-\log_{10}[\mathrm H^+], where [H+][\mathrm H^+] is the hydrogen-ion concentration in moles per litre. Because of the minus sign, a lower pH corresponds to a larger hydrogen-ion concentration. A difference of one pH unit represents a factor of 1010 in concentration.

The decibel model L=10log⁡10(II0)L=10\log_{10}(\frac{I}{I_0}) compares intensity II with a reference intensity I0I_0. The model matters: logarithms compare ratios, so the ratio must be positive and unitless. In any application, state what the model output represents rather than reporting an unexplained number.

Key terms

Logarithmic scale
A scale on which equal steps represent equal multiplicative factors.
Reference value
A fixed comparison quantity such as the intensity I0I_0 in a decibel model.
Order of magnitude
A factor of ten, represented by a difference of 1 in a base-10 logarithm.

Worked examples

Calculate pH

A solution has [H+]=3.2×10−5 mol L−1[\mathrm H^+]=3.2\times10^{-5}\text{ mol L}^{-1}. Find its pH.

  1. 1Use pH=−log⁡10(3.2×10−5)\mathrm{pH}=-\log_{10}(3.2\times10^{-5}).
  2. 2The calculator gives 4.494850…4.494850\ldots.
  3. 3Round as requested and interpret it as a pH value.

Answer: pH≈4.49\mathrm{pH}\approx4.49.

Compare sound intensities

A sound has intensity 250250 times a reference intensity. Find its level above the reference using L=10log⁡10(II0)L=10\log_{10}(\frac{I}{I_0}).

  1. 1Here II0=250\frac{I}{I_0}=250.
  2. 2Calculate L=10log⁡10(250)≈23.9794L=10\log_{10}(250)\approx23.9794.

Answer: The level is approximately 24.0 dB24.0\text{ dB} above the reference.

Interpret a magnitude difference

Two earthquake measurements are 6.26.2 and 4.74.7 on a simplified base-10 magnitude scale. Compare their measured amplitudes.

  1. 1The magnitude difference is 6.2−4.7=1.56.2-4.7=1.5.
  2. 2Convert the logarithmic difference to a ratio: 101.5≈31.622810^{1.5}\approx31.6228.

Answer: The larger measurement corresponds to about 31.631.6 times the measured amplitude in this simplified model.

Rules and formulas

pH model

pH=−log⁡10[H+]\mathrm{pH}=-\log_{10}[\mathrm H^+]

Sound-intensity level

L=10log⁡10(II0)L=10\log_{10}\left(\frac{I}{I_0}\right)

Ratio from a base-10 scale difference

ratio=10Δ\text{ratio}=10^{\Delta}
Δ\Delta
difference between the two scale readings

Calculator and GDC methods

Use the xyx^y, power, or caret key for general powers. Enter negative exponents with brackets when needed: type 5(−3)5^{(-3)}, not −53-5^3. Many calculators also distinguish the negative-number key from the subtraction key.

Use the log⁡\log key for base 1010 and the ln⁡\ln key for base ee. To evaluate another base, use change of base: log⁡ab=log⁡blog⁡a=ln⁡bln⁡a\log_a b=\frac{\log b}{\log a}=\frac{\ln b}{\ln a}. Never mix bases within the same ratio.

For a numerical exponential or logarithmic equation, a GDC graph or numerical solver can confirm the algebraic result. Define the two sides as functions and find their intersection, or enter the equation into the solver with a sensible initial estimate. In an IB solution, record the equation and relevant calculator result rather than writing only a decimal answer.

Retain guard digits. Store or reuse the full calculator value and round once at the end to the requested number of decimal places or significant figures.

For a model such as Q=Q0btQ=Q_0b^t, use a table to inspect whole-number times and a graph or solver for a continuous-time estimate. Set a window that includes both the model and the target line. An intersection gives the boundary value; the wording of the problem determines which side of that boundary is valid.

A reliable technology check uses two methods when possible. For example, solve 450(1.065)t=900450(1.065)^t=900 with t=ln⁡2ln⁡1.065t=\frac{\ln2}{\ln1.065}, then confirm by graphing y1=450(1.065)xy_1=450(1.065)^x and y2=900y_2=900. Agreement supports the calculation but does not replace units or interpretation.

Key terms

Change of base
A formula that evaluates a logarithm in any valid base using log⁡\log or ln⁡\ln.
Numerical solver
A GDC tool that approximates a solution to an entered equation.
Guard digits
Unrounded intermediate digits retained to protect final accuracy.
Intersection method
Graph both sides of an equation and find the input where their outputs are equal.

Worked examples

Evaluate a non-standard logarithm

Evaluate log⁡720\log_7 20 to 4 decimal places.

  1. 1Use change of base: log⁡720=ln⁡20ln⁡7\log_7 20=\frac{\ln20}{\ln7}.
  2. 2Enter the entire quotient with brackets if the calculator requires them.
  3. 3The display gives approximately 1.5395018491.539501849.
  4. 4Round once to 4 decimal places.

Answer: log⁡720≈1.5395\log_7 20\approx1.5395.

Confirm an equation graphically

Use a GDC to solve 3x=x+63^x=x+6 approximately.

  1. 1Enter Y1=3xY_1=3^x and Y2=x+6Y_2=x+6.
  2. 2Choose a window that displays both intersections.
  3. 3Use the intersection command near each crossing.
  4. 4The intersections occur at approximately x=−5.999x=-5.999 and x=1.879x=1.879.
  5. 5Substitute or inspect both sides to confirm each numerical result.

Answer: x≈−5.999x\approx-5.999 or x≈1.879x\approx1.879.

Use a table for a whole-period threshold

An account follows A=450(1.065)tA=450(1.065)^t. Find the first whole year in which it exceeds 900900.

  1. 1The boundary equation gives t=ln⁡2ln⁡1.065≈11.0067t=\frac{\ln2}{\ln1.065}\approx11.0067.
  2. 2A GDC table shows A(11)≈899.62<900A(11)\approx899.62<900.
  3. 3The next row gives A(12)>900A(12)>900.

Answer: The account first exceeds 900900 in year 1212.

Rules and formulas

Change-of-base formula

log⁡ab=log⁡blog⁡a=ln⁡bln⁡a\log_a b=\frac{\log b}{\log a}=\frac{\ln b}{\ln a}

IB-style practice with worked solutions

The following mixed questions bring together exponent laws, logarithmic notation, equation solving, domains, and calculator accuracy. Attempt each question without reading the steps first, then compare your working line by line.

In an IB response, exact values should remain exact unless a decimal is requested. For numerical answers, show the equation, keep full precision during working, and attach the requested accuracy to the final line.

Worked examples

Practice 1 · Product and quotient laws

Simplify x7x−2x3\frac{x^7x^{-2}}{x^3}, where x≠0x\ne0.

  1. 1Combine the numerator: x7+(−2)=x5x^{7+(-2)}=x^5.
  2. 2Divide: x5x3=x5−3\frac{x^5}{x^3}=x^{5-3}.

Answer: x2x^2.

Practice 2 · Zero and negative exponents

Evaluate 30+2−33^0+2^{-3}.

  1. 1Use 30=13^0=1.
  2. 2Use 2−3=123=182^{-3}=\frac{1}{2^3}=\frac{1}{8}.
  3. 3Add 1+181+\frac{1}{8}.

Answer: 98\frac{9}{8}.

Practice 3 · Simplification

Simplify (2a−1b2)34b−1\frac{(2a^{-1}b^2)^3}{4b^{-1}} using positive exponents.

  1. 1Cube the numerator: 8a−3b68a^{-3}b^6.
  2. 2Divide the coefficients: 84=2\frac{8}{4}=2.
  3. 3For bb, subtract −1-1: b6−(−1)=b7b^{6-(-1)}=b^7.
  4. 4Rewrite a−3a^{-3} in the denominator.

Answer: 2b7a3\frac{2b^7}{a^3}, for a,b≠0a,b\ne0.

Practice 4 · Convert notation

Write log⁡464=3\log_4 64=3 in exponential form and explain its meaning.

  1. 1The base is 44, the logarithm value is the exponent 33, and the argument is 6464.
  2. 2Use log⁡ab=c⟺ac=b\log_a b=c\Longleftrightarrow a^c=b.

Answer: 43=644^3=64; the exponent on 4 required to produce 64 is 3.

Practice 5 · Exact logarithm values

Evaluate log⁡2(116)+ln⁡1\log_2(\frac{1}{16})+\ln1.

  1. 1Since 116=2−4\frac{1}{16}=2^{-4}, log⁡2(116)=−4\log_2(\frac{1}{16})=-4.
  2. 2Since e0=1e^0=1, ln⁡1=0\ln1=0.
  3. 3Add the exact values.

Answer: −4-4.

Practice 6 · Exponential equation with a common base

Solve 9x−1=279^{x-1}=27.

  1. 1Rewrite with base 3: 9=329=3^2 and 27=3327=3^3.
  2. 2Then (32)x−1=32x−2=33(3^2)^{x-1}=3^{2x-2}=3^3.
  3. 3Equate exponents: 2x−2=32x-2=3.

Answer: x=52x=\frac{5}{2}.

Practice 7 · Numerical exponential equation

Solve 7x=507^x=50 to 3 significant figures.

  1. 1Take logarithms: xln⁡7=ln⁡50x\ln7=\ln50.
  2. 2Rearrange: x=ln⁡50ln⁡7x=\frac{\ln50}{\ln7}.
  3. 3Evaluate without intermediate rounding.

Answer: x≈2.01x\approx2.01 (3 s.f.).

Practice 8 · Logarithmic equation

Solve log⁡5(2x−1)=2\log_5(2x-1)=2.

  1. 1Require 2x−1>02x-1>0, so x>12x>\frac{1}{2}.
  2. 2Convert to exponential form: 2x−1=52=252x-1=5^2=25.
  3. 3Solve 2x=262x=26.
  4. 4Check that the argument is positive.

Answer: x=13x=13.

Practice 9 · Domain and extraneous roots

Solve ln⁡x+ln⁡(x−3)=ln⁡4\ln x+\ln(x-3)=\ln4.

  1. 1Require x>3x>3.
  2. 2Condense: ln⁡[x(x−3)]=ln⁡4\ln[x(x-3)]=\ln4.
  3. 3Set valid arguments equal: x(x−3)=4x(x-3)=4.
  4. 4Solve x2−3x−4=0x^2-3x-4=0, giving x=4x=4 or x=−1x=-1.
  5. 5Reject −1-1 because it violates x>3x>3.

Answer: x=4x=4 only.

Practice 10 · IB-style growth model

A quantity follows Q=80(1.12)tQ=80(1.12)^t. Find the first integer tt for which Q>150Q>150.

  1. 1Solve the boundary equation 80(1.12)t=15080(1.12)^t=150.
  2. 2Divide: (1.12)t=1.875(1.12)^t=1.875.
  3. 3Take logs: t=ln⁡1.875ln⁡1.12≈5.55t=\frac{\ln1.875}{\ln1.12}\approx5.55.
  4. 4Because tt must be an integer and the inequality is strict, test the next integer.

Answer: The first valid value is t=6t=6.

Practice 11 · Growth prediction

A wildlife population is modeled by P=1250(1.047)tP=1250(1.047)^t, where tt is in years. Find PP after 99 years to the nearest whole number.

  1. 1Substitute t=9t=9: P=1250(1.047)9P=1250(1.047)^9.
  2. 2The GDC gives P≈1889.862P\approx1889.862.
  3. 3Round the population to the nearest whole number.

Answer: Approximately 18901890 animals.

Practice 12 · Decay threshold

A pollutant level is modeled by C=72(0.91)tC=72(0.91)^t. Find the first whole day for which C<30C<30.

  1. 1Solve 72(0.91)t=3072(0.91)^t=30: t=ln⁡(3072)ln⁡0.91≈9.283t=\frac{\ln(\frac{30}{72})}{\ln0.91}\approx9.283.
  2. 2Because the model is decreasing, C<30C<30 for t>9.283t>9.283.
  3. 3The first allowable whole day is therefore 1010.

Answer: Day 1010.

Practice 13 · pH application

Find the pH when [H+]=6.4×10−4 mol L−1[\mathrm H^+]=6.4\times10^{-4}\text{ mol L}^{-1}.

  1. 1Use pH=−log⁡10(6.4×10−4)\mathrm{pH}=-\log_{10}(6.4\times10^{-4}).
  2. 2The GDC gives 3.193820…3.193820\ldots.

Answer: pH≈3.19\mathrm{pH}\approx3.19.

Practice 14 · Model interpretation

A social-media post is modeled by V=300(1.35)tV=300(1.35)^t, where tt is hours. A GDC gives t=8.79t=8.79 when V=4200V=4200. Explain two checks needed before reporting an answer.

  1. 1Check whether the context permits fractional hours or asks for the first complete hour.
  2. 2Check neighbouring values if a whole hour is required: compare V(8)V(8) and V(9)V(9) with 42004200.
  3. 3State that a constant 35% hourly growth rate may not remain realistic for long periods.

Answer: Report about 8.798.79 hours for continuous time, or hour 99 if the first qualifying whole hour is required, and state the constant-rate limitation.

Topic summary and cheat sheet

For exponents, identify the base and the operation before choosing a law. Products add same-base exponents, quotients subtract them, and a power of a power multiplies them. A zero exponent requires a non-zero base; a negative exponent means reciprocal.

For logarithms, translate the question into “what exponent?” Use ax=b⟺log⁡ab=xa^x=b\Longleftrightarrow\log_a b=x, with a>0a>0, a≠1a\ne1, and b>0b>0. Match bases for exact equations; otherwise isolate the power and use logarithms or technology.

For applications, write the model, identify units and time periods, calculate with guard digits, then interpret. A decimal time, a whole-period threshold and a long-term forecast are different kinds of answers.

Rules and formulas

Product and quotient laws

aman=am+n,aman=am−na^ma^n=a^{m+n},\qquad\frac{a^m}{a^n}=a^{m-n}

Power laws

(am)n=amn,(ab)n=anbn(a^m)^n=a^{mn},\qquad(ab)^n=a^nb^n

Zero and negative exponents

a0=1,a−n=1an(a≠0)a^0=1,\qquad a^{-n}=\frac1{a^n}\quad(a\ne0)

Exponential-logarithmic equivalence

ax=b⟺log⁡ab=xa^x=b\Longleftrightarrow\log_a b=x

Change of base

log⁡ab=log⁡blog⁡a=ln⁡bln⁡a\log_a b=\frac{\log b}{\log a}=\frac{\ln b}{\ln a}

Growth or decay model

Q(t)=Q0btQ(t)=Q_0b^t

Topic summary

  • For positive integers, ana^n represents repeated multiplication of the base aa.
  • For a≠0a\ne0, a0=1a^0=1 and a−n=1ana^{-n}=\frac{1}{a^n}.
  • Same-base multiplication adds exponents; same-base division subtracts them.
  • A power raised to a power multiplies the exponents: (am)n=amn(a^m)^n=a^{mn}.
  • Exponent laws act on products, quotients and powers—not on addition or subtraction.
  • The equivalence ax=b⟺log⁡ab=xa^x=b\Longleftrightarrow\log_a b=x defines a logarithm as an exponent.
  • For real logarithms, require a>0a>0, a≠1a\ne1, and a positive argument.
  • Use product, quotient and power laws only with valid positive logarithm arguments; there is no law for log⁡(x+y)\log(x+y).
  • Solve exponential equations by matching bases when possible; otherwise isolate the power and take logarithms.
  • Solve logarithmic equations by converting or condensing, then check candidates in the original equation.
  • Growth uses b=1+rb=1+r, decay uses b=1−rb=1-r, and Q=Q0btQ=Q_0b^t requires matching time periods.
  • To find time to a target, solve Q0bt=TQ_0b^t=T with logarithms and interpret any integer constraint.
  • Logarithmic scales turn multiplicative ratios into additive differences.
  • On a GDC, use brackets carefully, graph both sides or use a solver, and use a table for whole-period thresholds.
  • Retain guard digits and round only the final answer to the requested accuracy.
  • Every contextual answer should include units, a meaningful interpretation and any important model limitation.