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Mathematics AI · HL · Unit 1: Number and Algebra

Unit 1Topic 1.9Key Concepts69 min read

Laws of logarithms

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Meaning of logarithms and inverse relationships

A logarithm answers an exponent question. The statement log⁡ax=y\log_a x=y means that ay=xa^y=x. Exponential form and logarithmic form contain the same information.

For real logarithms, the base must satisfy a>0a>0 and a≠1a\ne1, while the argument must satisfy x>0x>0. These restrictions must be checked before and after solving equations.

The common logarithm log⁡x\log x usually means base 10, while ln⁡x\ln x means base ee, where e≈2.71828e\approx2.71828.

Two anchor facts follow directly from exponent rules: log⁡a1=0\log_a1=0 because a0=1a^0=1, and log⁡aa=1\log_a a=1 because a1=aa^1=a.

Key terms

Logarithm
The exponent to which a base must be raised to produce a given positive argument.
Base
The positive number a≠1a\ne1 in log⁡ax\log_a x.
Argument
The positive expression inside a logarithm.
Natural logarithm
The logarithm with base ee, written ln⁡\ln.

Worked examples

Convert to logarithmic form

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

  1. 1The base is 2, the exponent is 5, and the result is 32.

Answer: log⁡232=5\log_2 32=5.

Convert to exponential form

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

  1. 1Keep base 3, use exponent −3-3, and set the result equal to 127\frac{1}{27}.

Answer: 3−3=1273^{-3}=\frac{1}{27}.

Find the domain

Find the domain of f(x)=ln⁡(2x−6)f(x)=\ln(2x-6).

  1. 1The logarithm argument must be positive.
  2. 2Solve 2x−6>02x-6>0.
  3. 3Therefore x>3x>3.

Answer: The domain is (3,∞)(3,\infty).

Rules and formulas

Logarithm definition

log⁡ax=y⟺ay=x\log_a x=y\quad\Longleftrightarrow\quad a^y=x

Logarithm of one

log⁡a1=0\log_a1=0

Logarithm of its base

log⁡aa=1\log_a a=1

Inverse relationships

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

Why the logarithm laws work

The logarithm laws are exponent laws written in reverse. If u=log⁡axu=\log_a x and v=log⁡ayv=\log_a y, then x=aux=a^u and y=avy=a^v. Therefore xy=au+vxy=a^{u+v}, so log⁡a(xy)=u+v=log⁡ax+log⁡ay\log_a(xy)=u+v=\log_a x+\log_a y.

Similarly, xy=au−v\frac{x}{y}=a^{u-v} gives the quotient law. If x=aux=a^u, then xr=arux^r=a^{ru}, giving log⁡a(xr)=rlog⁡ax\log_a(x^r)=r\log_a x whenever the expressions are defined.

These derivations explain why multiplication becomes addition, division becomes subtraction, and an exponent becomes a coefficient. They also explain why there is no corresponding rule for log⁡a(x+y)\log_a(x+y): exponent laws contain no rule that turns a sum of powers into a single power.

Key terms

Inverse operation
An operation that reverses another operation; logarithms reverse exponentiation.
Derivation
A logical argument that obtains a rule from definitions and previously established results.
Common domain
The set of inputs for which every expression in an identity is defined.

Worked examples

Verify the product law numerically

Verify log⁡2(4×8)=log⁡24+log⁡28\log_2(4\times8)=\log_2 4+\log_2 8.

  1. 1The left side is log⁡232=5\log_2 32=5.
  2. 2The right side is 2+3=52+3=5.
  3. 3Both sides agree, illustrating the product law.

Answer: Both sides equal 55.

Derive a reciprocal identity

Show that log⁡a(1x)=−log⁡ax\log_a(\frac{1}{x})=-\log_a x for x>0x>0.

  1. 1Write 1x=x−1\frac{1}{x}=x^{-1}.
  2. 2Apply the power law: log⁡a(x−1)=−1log⁡ax\log_a(x^{-1})=-1\log_a x.

Answer: log⁡a(1x)=−log⁡ax\log_a(\frac{1}{x})=-\log_a x.

Rules and formulas

Product-law derivation

xy=auav=au+vxy=a^ua^v=a^{u+v}
uu
log⁡ax\log_a x
vv
log⁡ay\log_a y

Quotient-law derivation

xy=auav=au−v\frac{x}{y}=\frac{a^u}{a^v}=a^{u-v}

Product, quotient, and power laws of logarithms

Logarithm laws convert multiplication into addition, division into subtraction, and powers into coefficients. They follow from the corresponding exponent laws.

The product law applies to a product inside one logarithm. The quotient law applies to a quotient. The power law moves an exponent on the entire argument to the front.

There is no logarithm law for a sum or difference. Expressions such as log⁡a(x+y)\log_a(x+y) and log⁡a(x−y)\log_a(x-y) generally cannot be split.

Every separate logarithm created by expansion must have a positive argument. When variables appear, use the domain conditions supplied in the question or state appropriate restrictions.

Key terms

Product law
A logarithm of a product becomes the sum of logarithms.
Quotient law
A logarithm of a quotient becomes the difference of logarithms.
Power law
An exponent on a logarithm argument becomes a multiplier.
Identity
An equality true for every value in the common domain of both sides.

Worked examples

Apply all three laws

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

  1. 1Use the quotient law: log⁡2(8x3)−log⁡2(y12)\log_2(8x^3)-\log_2(y^{\frac{1}{2}}).
  2. 2Use product and power laws: log⁡28+3log⁡2x−12log⁡2y\log_2 8+3\log_2x-\frac12\log_2y.
  3. 3Since log⁡28=3\log_2 8=3, simplify.

Answer: 3+3log⁡2x−12log⁡2y3+3\log_2x-\frac12\log_2y.

Use the power law

Simplify log⁡5125x\log_5 125^x.

  1. 1Apply the power law: xlog⁡5125x\log_5 125.
  2. 2Since 125=53125=5^3, log⁡5125=3\log_5 125=3.

Answer: 3x3x.

Recognize an invalid split

Is ln⁡(x+4)=ln⁡x+ln⁡4\ln(x+4)=\ln x+\ln4 an identity?

  1. 1The product law applies to ln⁡(4x)\ln(4x), not ln⁡(x+4)\ln(x+4).
  2. 2Test x=1x=1: ln⁡5≠ln⁡1+ln⁡4=ln⁡4\ln5\ne\ln1+\ln4=\ln4.

Answer: No. Logarithms do not split across addition.

Rules and formulas

Product law

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

Quotient law

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

Power law

log⁡a(xr)=rlog⁡ax\log_a(x^r)=r\log_a x

Reciprocal result

log⁡a(1x)=−log⁡ax\log_a\left(\frac1x\right)=-\log_a x

Expanding, condensing, and changing the base

Expanding rewrites one logarithm as several simpler terms. Condensing reverses the process and combines several logarithms into one. Coefficients must become exponents before product or quotient laws are used.

The change-of-base formula lets a GDC evaluate logarithms in any valid base using log⁡\log or ln⁡\ln. Both numerator and denominator must use the same new base.

Calculator answers are approximations unless the logarithm simplifies exactly. Retain full precision internally and round only at the final step.

Key terms

Expand
Rewrite a logarithm of products, quotients, or powers as a sum or difference.
Condense
Combine logarithmic terms into a single logarithm.
Change of base
Rewrite a logarithm using another valid base, usually 1010 or ee, for evaluation.
GDC
A graphical display calculator used to evaluate, graph, and solve expressions numerically.

Worked examples

Condense a logarithmic expression

Condense 2ln⁡x−12ln⁡y+ln⁡32\ln x-\frac12\ln y+\ln3, where x,y>0x,y>0.

  1. 1Use the power law backwards: ln⁡(x2)−ln⁡(y12)+ln⁡3\ln(x^2)-\ln(y^{\frac{1}{2}})+\ln3.
  2. 2Combine addition as multiplication and subtraction as division.

Answer: ln⁡(3x2y)\ln\left(\frac{3x^2}{\sqrt y}\right).

Evaluate by change of base

Evaluate log⁡720\log_7 20 to three significant figures.

  1. 1Enter ln⁡20ln⁡7\frac{\ln20}{\ln7} or log⁡20log⁡7\frac{\log20}{\log7}.
  2. 2The calculator gives approximately 1.539501849561.53950184956.
  3. 3Round to three significant figures.

Answer: log⁡720≈1.54\log_7 20\approx1.54.

Expand before substituting

Given log⁡a2=p\log_a2=p and log⁡a3=q\log_a3=q, express log⁡a72\log_a72 in terms of p,qp,q.

  1. 1Factor 72=23⋅3272=2^3\cdot3^2.
  2. 2Apply product and power laws.
  3. 3log⁡a72=3log⁡a2+2log⁡a3\log_a72=3\log_a2+2\log_a3.

Answer: 3p+2q3p+2q.

Rules and formulas

Change of base

log⁡ax=log⁡bxlog⁡ba\log_a x=\frac{\log_b x}{\log_b a}

Calculator forms

log⁡ax=ln⁡xln⁡a=log⁡xlog⁡a\log_a x=\frac{\ln x}{\ln a}=\frac{\log x}{\log a}

Condensing pattern

rlog⁡ax−slog⁡ay=log⁡a(xrys)r\log_a x-s\log_a y=\log_a\left(\frac{x^r}{y^s}\right)

Domain restrictions and advanced identity checks

For a real logarithm log⁡a(g(x))\log_a(g(x)), require a>0a>0, a≠1a\ne1, and g(x)>0g(x)>0. If an equation contains several logarithms, every argument must be positive simultaneously.

Restrictions should be written before logarithms are combined or removed. Algebra may produce candidates outside the original domain, and those candidates must be rejected by substitution into the original equation.

The familiar power law log⁡a(xr)=rlog⁡ax\log_a(x^r)=r\log_a x is stated with x>0x>0. A useful HL refinement is log⁡a(x2)=2log⁡a∣x∣\log_a(x^2)=2\log_a|x| for x≠0x\ne0; writing 2log⁡ax2\log_a x would incorrectly exclude negative xx.

A base between 00 and 11 is valid. In this case log⁡ax\log_a x is decreasing, so inequalities reverse when converting between logarithmic and exponential forms.

Key terms

Domain restriction
A condition that keeps every logarithm argument positive and every base valid.
Extraneous root
An algebraic candidate that fails the original equation or its domain conditions.
Decreasing logarithm
A logarithmic function with base 0<a<10<a<1, whose output decreases as its input increases.

Worked examples

Find a compound logarithmic domain

Find the domain of f(x)=ln⁡(x−2)+log⁡3(7−x)f(x)=\ln(x-2)+\log_3(7-x).

  1. 1Require x−2>0x-2>0, so x>2x>2.
  2. 2Require 7−x>07-x>0, so x<7x<7.
  3. 3Intersect the two conditions.

Answer: The domain is (2,7)(2,7).

Use absolute value correctly

Simplify log⁡5(x2)\log_5(x^2) for real x≠0x\ne0.

  1. 1Since x2=∣x∣2x^2=|x|^2 and ∣x∣>0|x|>0, apply the power law to ∣x∣|x|.

Answer: log⁡5(x2)=2log⁡5∣x∣\log_5(x^2)=2\log_5|x|.

Solve an inequality with a decreasing base

Solve log⁡12x>2\log_{\frac{1}{2}}x>2.

  1. 1The domain is x>0x>0.
  2. 2Because 0<12<10<\frac{1}{2}<1, the logarithm is decreasing, so the inequality reverses when converted.
  3. 3Thus x<(12)2=14x<(\frac{1}{2})^2=\frac{1}{4}.

Answer: 0<x<140<x<\frac14.

Rules and formulas

Real logarithm conditions

a>0,a≠1,g(x)>0a>0,\qquad a\ne1,\qquad g(x)>0

Power identity for unrestricted nonzero x

log⁡a(x2)=2log⁡a∣x∣\log_a(x^2)=2\log_a|x|

Solving exponential and logarithmic equations

For exponential equations, first try to express both sides with the same base. If that is impossible, isolate the exponential expression and take logarithms.

For logarithmic equations, state domain restrictions, combine logarithms when useful, convert to exponential form, solve the resulting equation, and check every candidate in the original equation.

Taking logarithms is valid only after the isolated exponential side is positive. This is automatically true for af(x)a^{f(x)} when a>0a>0, but coefficients and other terms still need careful algebra.

A GDC can solve equations numerically using a solver or graph intersections. An IB response should still show the equation entered, the chosen interval or graph, and suitable rounding.

Key terms

Extraneous solution
A value produced during algebra that does not satisfy the original equation or its domain.
Same-base method
Rewrite both sides with one base and equate exponents.
Logarithmic method
Take logarithms after isolating an exponential expression.
Intersection method
Graph each side as a function and read the coordinates where the graphs meet.

Worked examples

Solve using a common base

Solve 4x=8x−14^x=8^{x-1}.

  1. 1Rewrite using base 2: 22x=23x−32^{2x}=2^{3x-3}.
  2. 2Equate exponents: 2x=3x−32x=3x-3.
  3. 3Solve to obtain x=3x=3.

Answer: x=3x=3.

Solve by taking logarithms

Solve 32x−1=203^{2x-1}=20.

  1. 1Take natural logarithms: (2x−1)ln⁡3=ln⁡20(2x-1)\ln3=\ln20.
  2. 2Rearrange: 2x−1=ln⁡20ln⁡32x-1=\frac{\ln20}{\ln3}.
  3. 3Therefore x=12(1+ln⁡20ln⁡3)x=\frac12\left(1+\frac{\ln20}{\ln3}\right).

Answer: x≈1.86x\approx1.86 to three significant figures.

Solve and reject an invalid root

Solve log⁡2x+log⁡2(x−2)=3\log_2x+\log_2(x-2)=3.

  1. 1Domain: x>2x>2.
  2. 2Combine: log⁡2[x(x−2)]=3\log_2[x(x-2)]=3.
  3. 3Convert: x(x−2)=8x(x-2)=8, so x2−2x−8=0x^2-2x-8=0.
  4. 4The candidates are x=4x=4 and x=−2x=-2; reject −2-2 because it violates the domain.

Answer: x=4x=4.

Solve a growth equation

Solve 2e0.4t=152e^{0.4t}=15.

  1. 1Isolate the exponential: e0.4t=7.5e^{0.4t}=7.5.
  2. 2Take natural logarithms: 0.4t=ln⁡7.50.4t=\ln7.5.
  3. 3Divide by 0.40.4.

Answer: t=ln⁡7.50.4≈5.04t=\frac{\ln7.5}{0.4}\approx5.04.

Rules and formulas

Same-base principle

au=av⟹u=va^u=a^v\quad\Longrightarrow\quad u=v

Logarithmic solution step

af(x)=k⟹f(x)=ln⁡kln⁡aa^{f(x)}=k\quad\Longrightarrow\quad f(x)=\frac{\ln k}{\ln a}

Log equality principle

log⁡au=log⁡av⟹u=v\log_a u=\log_a v\quad\Longrightarrow\quad u=v

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

Logarithmic models and linearization

Logarithms can transform some nonlinear models into straight-line form. For y=Abxy=Ab^x, taking natural logarithms gives ln⁡y=ln⁡A+xln⁡b\ln y=\ln A+x\ln b. A plot of ln⁡y\ln y against xx is linear with gradient ln⁡b\ln b and vertical intercept ln⁡A\ln A.

For a power model y=Axny=Ax^n, taking logarithms gives ln⁡y=ln⁡A+nln⁡x\ln y=\ln A+n\ln x. A plot of ln⁡y\ln y against ln⁡x\ln x is linear with gradient nn and intercept ln⁡A\ln A. Common logarithms work equally well if the same base is used throughout.

After fitting a transformed line, convert the intercept back to the original parameter: if the intercept is cc, then A=ecA=e^c for natural logs or A=10cA=10^c for common logs. Interpret the gradient and intercept in the original context.

A strong model check examines the transformed scatter plot and residuals, then considers whether constant percentage growth or a power relationship is reasonable. A high correlation does not prove the model will remain reliable beyond the observed data.

Key terms

Linearization
Transforming variables so that a nonlinear relationship becomes a straight-line relationship.
Semi-log plot
A plot using one logarithmic variable, such as ln⁡y\ln y against xx.
Log-log plot
A plot of ln⁡y\ln y against ln⁡x\ln x, used to investigate a power model.
Back-transform
Undo a logarithm to recover a parameter or prediction in the original units.

Worked examples

Linearize an exponential model

Write P=320(1.07)tP=320(1.07)^t in straight-line form using Y=ln⁡PY=\ln P.

  1. 1Take natural logarithms: ln⁡P=ln⁡[320(1.07)t]\ln P=\ln[320(1.07)^t].
  2. 2Apply product and power laws: ln⁡P=ln⁡320+tln⁡1.07\ln P=\ln320+t\ln1.07.
  3. 3With Y=ln⁡PY=\ln P, the gradient is ln⁡1.07\ln1.07 and the intercept is ln⁡320\ln320.

Answer: Y=(ln⁡1.07)t+ln⁡320Y=(\ln1.07)t+\ln320.

Recover a power model

A regression gives log⁡10y=0.699+1.8log⁡10x\log_{10}y=0.699+1.8\log_{10}x. Find the power model.

  1. 1Compare with log⁡10y=log⁡10A+nlog⁡10x\log_{10}y=\log_{10}A+n\log_{10}x.
  2. 2The exponent is n=1.8n=1.8.
  3. 3The intercept gives log⁡10A=0.699\log_{10}A=0.699, so A=100.699≈5.00A=10^{0.699}\approx5.00.

Answer: y≈5.00x1.8y\approx5.00x^{1.8}.

Interpret a transformed gradient

A plot of ln⁡M\ln M against tt has gradient −0.23-0.23. Interpret the multiplier per unit time.

  1. 1For M=AektM=Ae^{kt}, the transformed gradient is kk.
  2. 2Here k=−0.23k=-0.23, so the multiplier is e−0.23≈0.795e^{-0.23}\approx0.795.
  3. 3The corresponding percentage decrease is about (1−0.795)×100%=20.5%(1-0.795)\times100\%=20.5\%.

Answer: The model multiplies by about 0.795 per unit time, a decrease of about 20.5% per unit time.

Rules and formulas

Exponential linearization

y=Abx⟹ln⁡y=ln⁡A+xln⁡by=Ab^x\quad\Longrightarrow\quad \ln y=\ln A+x\ln b

Power-model linearization

y=Axn⟹ln⁡y=ln⁡A+nln⁡xy=Ax^n\quad\Longrightarrow\quad \ln y=\ln A+n\ln x

Back-transform an intercept

c=ln⁡A⟹A=ecc=\ln A\quad\Longrightarrow\quad A=e^c

Calculator and GDC methods

Use log⁡\log for base 1010 and ln⁡\ln for base ee. For any other valid base, enter log⁡ax=ln⁡xln⁡a\log_a x=\frac{\ln x}{\ln a} or use the calculator's dedicated logarithm-base template.

For a numerical equation, graph the two sides and find every intersection in the stated domain, or use a numerical solver with sensible starting values. Scan the graph first because one initial estimate may locate only one of several roots.

When working with transformed data, create lists for the original variables and calculated lists for their logarithms. Run the linear regression on the transformed lists, record the regression equation, and back-transform its intercept before writing the original model.

Keep guard digits and round once at the end. In an IB response, write the equation or regression entered, report relevant coordinates or parameters, include units, and interpret the result in context.

Key terms

Numerical solver
A GDC tool that approximates roots of an entered equation.
Intersection method
Graphing each side of an equation and locating points where their outputs are equal.
Guard digits
Unrounded intermediate digits retained until the final answer.
Transformed regression
A regression performed on logarithms of one or both variables to fit a nonlinear model.

Worked examples

Evaluate a logarithm with change of base

Evaluate log⁡735\log_7 35 to four decimal places.

  1. 1Enter (ln⁡35)(ln⁡7)\frac{(\ln35)}{(\ln7)}.
  2. 2The display gives 1.827087…1.827087\ldots.
  3. 3Round once to four decimal places.

Answer: log⁡735≈1.8271\log_7 35\approx1.8271.

Find all roots graphically

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

  1. 1Enter Y1=3xY_1=3^x and Y2=x+6Y_2=x+6.
  2. 2Choose a wide enough window to display both crossings.
  3. 3Use the intersection command near each crossing and verify the two outputs agree.

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

Check a whole-period threshold

An account follows A=450(1.065)tA=450(1.065)^t. Find the first whole year in which A>900A>900.

  1. 1Solve the boundary equation to obtain t=ln⁡2ln⁡1.065≈11.0067t=\frac{\ln2}{\ln1.065}\approx11.0067.
  2. 2Use a table: A(11)≈899.62<900A(11)\approx899.62<900, while A(12)>900A(12)>900.

Answer: The account first exceeds 900 in year 12.

Rules and formulas

Change of base for a GDC

log⁡ax=ln⁡xln⁡a=log⁡xlog⁡a\log_a x=\frac{\ln x}{\ln a}=\frac{\log x}{\log a}

IB AI HL-style practice and exam strategy

These questions mix exact algebra, numerical technology, modelling, and interpretation. Show domain restrictions and exact working before using decimal approximations.

For application questions, identify the model and time unit, keep full calculator precision, apply any integer or domain constraint, and finish with a sentence that answers the question in context.

Key terms

Exact form
An unrounded expression such as a logarithmic quotient or radical.
Admissible solution
A solution that satisfies the original equation, its domain, and any contextual constraints.
Model validity
The range and circumstances in which a model's assumptions remain reasonable.

Worked examples

Practice 1 · Exponential and logarithmic form

Write log⁡5125=3\log_5 125=3 in exponential form.

  1. 1Keep the base 5, use exponent 3, and set the result equal to 125.

Answer: 53=1255^3=125.

Practice 2 · Exact evaluation

Evaluate log⁡128\log_{\frac{1}{2}}8.

  1. 1Solve (12)y=8=23(\frac{1}{2})^y=8=2^3.
  2. 2Since (12)y=2−y(\frac{1}{2})^y=2^{-y}, −y=3-y=3.

Answer: −3-3.

Practice 3 · Expand logarithms

Expand log⁡3(27x4y)\log_3\left(\frac{27x^4}{\sqrt y}\right), where x,y>0x,y>0.

  1. 1Use quotient, product, and power laws.
  2. 2log⁡327+4log⁡3x−12log⁡3y\log_3 27+4\log_3x-\frac12\log_3y.
  3. 3Use log⁡327=3\log_3 27=3.

Answer: 3+4log⁡3x−12log⁡3y3+4\log_3x-\frac12\log_3y.

Practice 4 · Condense logarithms

Condense 2ln⁡x−3ln⁡y+12ln⁡z2\ln x-3\ln y+\frac12\ln z, where x,y,z>0x,y,z>0.

  1. 1Move coefficients to powers, combine added terms as products, and subtracted terms as quotients.

Answer: ln⁡(x2zy3)\ln\left(\frac{x^2\sqrt z}{y^3}\right).

Practice 5 · Change of base

Evaluate log⁡735\log_7 35 to three significant figures.

  1. 1Use log⁡735=ln⁡35ln⁡7\log_7 35=\frac{\ln35}{\ln7}.
  2. 2The GDC gives 1.827087…1.827087\ldots.

Answer: 1.831.83 to three significant figures.

Practice 6 · Single logarithm equation

Solve log⁡4(2x−1)=3\log_4(2x-1)=3.

  1. 1Domain: 2x−1>02x-1>0.
  2. 2Convert to exponential form: 2x−1=43=642x-1=4^3=64.
  3. 3Solve and check the domain.

Answer: x=652x=\frac{65}{2}.

Practice 7 · Domain rejection

Solve ln⁡(x−2)+ln⁡(x+1)=ln⁡10\ln(x-2)+\ln(x+1)=\ln10.

  1. 1Domain: x>2x>2.
  2. 2Combine: (x−2)(x+1)=10(x-2)(x+1)=10.
  3. 3Solve x2−x−12=0x^2-x-12=0, giving x=4x=4 or x=−3x=-3.
  4. 4Reject −3-3.

Answer: x=4x=4.

Practice 8 · Exponential threshold

Solve 7(1.12)t=207(1.12)^t=20 and interpret tt as time.

  1. 1Isolate the power: (1.12)t=207(1.12)^t=\frac{20}{7}.
  2. 2Take logarithms: t=ln⁡(207)ln⁡1.12t=\frac{\ln(\frac{20}{7})}{\ln1.12}.

Answer: t≈9.26t\approx9.26 time units.

Practice 9 · Logarithmic scale

Find the pH when [H+]=2.5×10−6 mol L−1[\mathrm H^+]=2.5\times10^{-6}\text{ mol L}^{-1}.

  1. 1Use pH=−log⁡10(2.5×10−6)\mathrm{pH}=-\log_{10}(2.5\times10^{-6}).
  2. 2Evaluate and round appropriately.

Answer: pH≈5.60\mathrm{pH}\approx5.60.

Practice 10 · Recover a power model

A transformed regression is ln⁡y=1.4+0.75ln⁡x\ln y=1.4+0.75\ln x. Find y=Axny=Ax^n.

  1. 1The gradient gives n=0.75n=0.75.
  2. 2The intercept gives ln⁡A=1.4\ln A=1.4, so A=e1.4≈4.055A=e^{1.4}\approx4.055.

Answer: y≈4.06x0.75y\approx4.06x^{0.75}.

Practice 11 · Numerical equation

Use a GDC to solve e−x=xe^{-x}=x.

  1. 1Graph y1=e−xy_1=e^{-x} and y2=xy_2=x.
  2. 2Find their intersection and verify both outputs agree.

Answer: x≈0.567x\approx0.567.

Practice 12 · Domain of a composite argument

Find the domain of f(x)=log⁡2(x2−9)f(x)=\log_2(x^2-9).

  1. 1Require x2−9>0x^2-9>0.
  2. 2Factor: (x−3)(x+3)>0(x-3)(x+3)>0.
  3. 3Use a sign diagram or interval reasoning.

Answer: (−∞,−3)∪(3,∞)( -\infty,-3)\cup(3,\infty).

Rules and formulas

Equation workflow

DOMAIN→TRANSFORM→SOLVE→CHECK→INTERPRET\text{DOMAIN}\to\text{TRANSFORM}\to\text{SOLVE}\to\text{CHECK}\to\text{INTERPRET}

Exponential threshold time

t=ln⁡(TQ0)ln⁡bt=\frac{\ln(\frac{T}{Q_0})}{\ln b}

Topic summary and cheat sheet

Use this checklist to choose a method quickly. First identify the logarithm base and all argument restrictions; then decide whether the task needs conversion, a law, an equation method, a model, or technology.

Exact algebra and contextual interpretation are equally important. A correct calculator value is not a complete IB AI HL answer unless its setup, accuracy, units, constraints, and meaning are clear.

Key terms

Product
Inside multiplication becomes outside addition.
Quotient
Inside division becomes outside subtraction.
Power
An exponent on the argument becomes a coefficient.
Change of base
Use one consistent new base in numerator and denominator.

Worked examples

Rapid method choice

Choose the first useful step for 4e0.3x=114e^{0.3x}=11.

  1. 1Isolate the exponential: e0.3x=114e^{0.3x}=\frac{11}{4}.
  2. 2Then take natural logarithms.

Answer: x=ln⁡(114)0.3x=\frac{\ln(\frac{11}{4})}{0.3}.

Rapid domain check

State the restriction for log⁡5(3−2x)\log_5(3-2x).

  1. 1Require 3−2x>03-2x>0.

Answer: x<32x<\frac32.

Rules and formulas

Definition

log⁡ax=y  ⟺  ay=x\log_a x=y\iff a^y=x

Product law

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

Quotient law

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

Power law

log⁡a(xr)=rlog⁡ax\log_a(x^r)=r\log_a x

Change of base

log⁡ax=ln⁡xln⁡a\log_a x=\frac{\ln x}{\ln a}

Valid real logarithm

a>0, a≠1, x>0a>0,\ a\ne1,\ x>0

Topic summary

  • log⁡ax=y  ⟺  ay=x\log_a x=y\iff a^y=x, with a>0a>0, a≠1a\ne1, and x>0x>0.
  • log⁡a1=0\log_a1=0, log⁡aa=1\log_a a=1, and logarithms reverse exponentiation.
  • log⁡a(xy)=log⁡ax+log⁡ay\log_a(xy)=\log_a x+\log_a y.
  • log⁡a(xy)=log⁡ax−log⁡ay\log_a(\frac{x}{y})=\log_a x-\log_a y.
  • log⁡a(xr)=rlog⁡ax\log_a(x^r)=r\log_a x when the argument conditions are satisfied.
  • There is no logarithm law for a sum or difference.
  • Change base with log⁡ax=ln⁡xln⁡a\log_a x=\frac{\ln x}{\ln a}; use the same new base in numerator and denominator.
  • To condense, move coefficients to exponents before combining sums as products and differences as quotients.
  • For equations use DOMAIN → TRANSFORM → SOLVE → CHECK → INTERPRET.
  • Rewrite exponential equations with a common base where possible; otherwise isolate the power and take logarithms.
  • In Q=Q0btQ=Q_0b^t, a target time is t=ln⁡(TQ0)ln⁡bt=\frac{\ln(\frac{T}{Q_0})}{\ln b}.
  • Logarithmic-scale differences represent multiplicative ratios, not ordinary additive changes.
  • For y=Abxy=Ab^x, plot ln⁡y\ln y against xx: gradient =ln⁡b=\ln b, intercept =ln⁡A=\ln A.
  • For y=Axny=Ax^n, plot ln⁡y\ln y against ln⁡x\ln x: gradient =n=n, intercept =ln⁡A=\ln A.
  • A GDC result must be accompanied by the entered equation or regression, appropriate accuracy, and contextual interpretation.
  • Keep guard digits, check every admissible solution, and apply whole-period constraints only after solving the continuous boundary.