Laws of logarithms AI HL

IB Mathematics: Applications and Interpretation · AI HL

Lesson 1.9 · Laws of Logarithms

Learn to expand, combine and solve logarithmic expressions, then apply them to real situations.

Faith connection · God’s love

Jesus said, “For God so loved the world, that he gave his only begotten Son” (John 3:16, KJV). Logarithms help us understand quantities across very different scales. God’s love cannot be measured by a formula, but He has shown it through the gift of His Son. Come to Jesus, trust Him as your Savior, and follow Him as Lord.

Learning objectives

By the end of this lesson, you should be able to:

use the product, quotient and power laws of logarithms;
expand logarithmic expressions and combine them into a single logarithm;
solve logarithmic equations and reject values outside their domains;
apply logarithmic laws to pH, sound intensity and growth models.

1 · A reminder: what does a logarithm mean?

A logarithm gives the exponent needed to obtain a number. The base $a$ must be positive and different from 1, and the argument $x$ must be positive.

$\log_a x=y\quad\Longleftrightarrow\quad a^y=x,\qquad a>0,\ a\ne1,\ x>0.$

Unless another base is written, $\log x$ means $\log_{10}x$ . Also, $\ln x$ means $\log_e x$ .

Worked example

Find $\log_2 32$ .

$2^5=32\quad\Longrightarrow\quad\log_2 32=5.$

2 · Product law

The logarithm of a product becomes a sum of logarithms. Here $x>0$ and $y>0$ .

$\boxed{\log_a(xy)=\log_a x+\log_a y}$

Why does this work? If $x=a^m$ and $y=a^n$ , then $xy=a^{m+n}$ .

Worked example

Evaluate $\log 8+\log 125$ without a calculator.

$\log 8+\log 125=\log(8\cdot125)=\log1000=\boxed{3}.$

Take care: The product law does not apply to a sum.

$\log_a(x+y)\ne\log_a x+\log_a y\quad\text{in general.}$

3 · Quotient law

The logarithm of a quotient becomes a difference of logarithms.

$\boxed{\log_a\left(\frac{x}{y}\right)=\log_a x-\log_a y},\qquad x>0,\ y>0.$

Worked example

Evaluate $\log_2 48-\log_2 3$ .

$\log_2 48-\log_2 3 =\log_2\left(\frac{48}{3}\right) =\log_2 16 =\boxed{4}.$

The order matters: the argument of the logarithm being subtracted goes in the denominator.

4 · Power law

An exponent inside a logarithm becomes a multiplier outside it.

$\boxed{\log_a(x^r)=r\log_a x},\qquad x>0.$

Worked example

Expand the following expression, assuming $x>0$ and $y>0$ .

$\ln\left(\frac{x^3\sqrt y}{5}\right)$

Use the quotient law, followed by the product and power laws.

$\ln\left(\frac{x^3\sqrt y}{5}\right) =\boxed{3\ln x+\frac12\ln y-\ln5}.$

Domain detail: If $x$ could be negative, then $\ln(x^2)=2\ln|x|$ for $x\ne0$ . Writing $2\ln x$ would wrongly exclude negative values.

5 · Solving logarithmic equations

Start by finding the values for which every logarithm in the original equation is defined. Then combine logarithms, solve and check your answer.

Worked example A · Product law

Solve the equation below.

$\log_3(x-1)+\log_3(x+1)=2.$

Both arguments must be positive, so $x>1$ . Combine the logarithms.

$\log_3\bigl((x-1)(x+1)\bigr)=2.$

Rewrite in exponential form and solve.

$x^2-1=3^2=9,\qquad x=\pm\sqrt{10}.$

Only the positive value satisfies $x>1$ .

$\boxed{x=\sqrt{10}}$

Worked example B · Quotient law

Solve the equation below.

$\log_2(x+2)-\log_2(x-1)=2.$

The domain requires $x>1$ . Combine the logarithms and solve.

$\log_2\left(\frac{x+2}{x-1}\right)=2.$

$\frac{x+2}{x-1}=4 \quad\Longrightarrow\quad x+2=4x-4 \quad\Longrightarrow\quad\boxed{x=2}.$

The solution satisfies $x>1$ .

IB-style Paper 1 · Skill practice

Show your working and check the domain of each equation.

1. [3 marks]

Given $\log_3 2=p$ and $\log_3 5=q$ , express $\log_3 200$ in terms of $p$ and $q$ .

Show solution

$200=2^3\cdot5^2.$

$\log_3 200=3\log_3 2+2\log_3 5=\boxed{3p+2q}.$

2. [4 marks]

Expand the expression below fully, where $x>0$ and $y>0$ .

$\log\left(\frac{100x^2}{\sqrt y}\right)$

Show solution

$\log\left(\frac{100x^2}{\sqrt y}\right) =\log100+\log(x^2)-\log(y^{1/2}).$

$\boxed{2+2\log x-\frac12\log y}$

3. [3 marks]

Write the expression below as a single logarithm, where $x>0$ .

$3\ln x-\frac12\ln(x+4)+\ln2$

Show solution

$3\ln x-\frac12\ln(x+4)+\ln2 =\ln(x^3)-\ln\sqrt{x+4}+\ln2.$

$\boxed{\ln\left(\frac{2x^3}{\sqrt{x+4}}\right)}$

4. [5 marks]

Solve the equation below. Give an exact answer.

$2\log x-\log4=1.$

Show solution

The domain requires $x>0$ .

$\log\left(\frac{x^2}{4}\right)=1 \quad\Longrightarrow\quad\frac{x^2}{4}=10.$

$x^2=40\quad\Longrightarrow\quad\boxed{x=2\sqrt{10}}.$

The negative root is outside the domain.

5. [5 marks]

Solve the equation below.

$\ln(x+3)-\ln(x-1)=\ln3.$

Show solution

The domain requires $x>1$ .

$\ln\left(\frac{x+3}{x-1}\right)=\ln3 \quad\Longrightarrow\quad\frac{x+3}{x-1}=3.$

$x+3=3x-3 \quad\Longrightarrow\quad\boxed{x=3}.$

6. [6 marks]

Find every real solution of the equation below.

$\log(x^2-9)=1+\log(x-3).$

Show solution

The original equation requires $x>3$ . Since $1=\log10$ :

$\log(x^2-9)=\log\bigl(10(x-3)\bigr).$

$(x-3)(x+3)=10(x-3).$

Because $x>3$ , divide by $x-3$ .

$x+3=10\quad\Longrightarrow\quad\boxed{x=7}.$

IB-style applications · Advanced practice

Application 1 · Acidity [6 marks]

The pH of a solution is given by the model below, where $[\mathrm{H}^+]$ is the hydrogen ion concentration in mol dm $^{-3}$ .

$\mathrm{pH}=-\log[\mathrm{H}^+].$

Solution A has pH 3.2 and solution B has pH 5.6. Find the ratio of the hydrogen ion concentration in A to that in B. Give your answer to three significant figures.

Show solution

$[\mathrm{H}^+]_{\mathrm A}=10^{-3.2}, \qquad [\mathrm{H}^+]_{\mathrm B}=10^{-5.6}.$

$\frac{[\mathrm{H}^+]_{\mathrm A}} {[\mathrm{H}^+]_{\mathrm B}} =\frac{10^{-3.2}}{10^{-5.6}} =10^{2.4} \approx\boxed{251}.$

Solution A has about 251 times the hydrogen ion concentration of solution B.

Application 2 · Sound intensity [6 marks]

The sound level $L$ , in decibels, is modeled by the equation below. Here $I$ is sound intensity and $I_0$ is a reference intensity.

$L=10\log\left(\frac{I}{I_0}\right).$

One sound has level 72 dB and another has level 58 dB. Find the ratio of their intensities to three significant figures.

Show solution

Call the intensities $I_1$ and $I_2$ . Subtract the sound levels and apply the quotient law.

$72-58 =10\log\left(\frac{I_1}{I_0}\right) -10\log\left(\frac{I_2}{I_0}\right).$

$14=10\log\left(\frac{I_1}{I_2}\right) \quad\Longrightarrow\quad \log\left(\frac{I_1}{I_2}\right)=1.4.$

$\boxed{\frac{I_1}{I_2}=10^{1.4}\approx25.1}.$

Application 3 · Exponential growth [7 marks]

A culture initially contains 500 bacteria. Its population after $t$ hours is modeled by:

$N(t)=500e^{kt}.$

After 3 hours, the culture contains 900 bacteria. (a) Find an exact expression for $k$ . (b) Predict the population after 8 hours, rounded to the nearest whole number.

Show solution

(a) Substitute $t=3$ .

$900=500e^{3k} \quad\Longrightarrow\quad e^{3k}=1.8.$

$\boxed{k=\frac{\ln1.8}{3}}.$

(b) Substitute into the model.

$N(8)=500e^{(8/3)\ln1.8} =500(1.8)^{8/3} \approx\boxed{2396}.$

Application 4 · Dose and response [7 marks]

A positive dose $d$ produces the modeled response:

$R=4+3\log d.$

(a) Find the increase in $R$ when the dose is multiplied by 20. (b) Find the factor by which the dose must be multiplied to increase $R$ by 5. Give numerical answers to three significant figures.

Show solution

(a) Use the product law.

$\Delta R=3\log(20d)-3\log d =3\log20 \approx\boxed{3.90}.$

(b) Let the multiplication factor be $f$ .

$3\log f=5 \quad\Longrightarrow\quad\log f=\frac53.$

$\boxed{f=10^{5/3}\approx46.4}.$

Quick recap

For positive $x$ and $y$ , and any valid base $a$ :

$\log_a(xy)=\log_a x+\log_a y$

$\log_a\left(\frac{x}{y}\right)=\log_a x-\log_a y$

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

Exam habit: Check that every logarithm in the original equation has a positive argument.

Lesson 1.9 · Laws of Logarithms

Faith connection · God’s love

Learning objectives

1 · A reminder: what does a logarithm mean?

2 · Product law

3 · Quotient law

4 · Power law

5 · Solving logarithmic equations

IB-style Paper 1 · Skill practice

IB-style applications · Advanced practice

Quick recap

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IB Mathematics: Applications and Interpretation HL 6 Sections80 LessonsLifetime

Curriculum

Laws of logarithms AI HL

Faith connection · God’s love

Learning objectives

1 · A reminder: what does a logarithm mean?

2 · Product law

3 · Quotient law

4 · Power law

5 · Solving logarithmic equations

IB-style Paper 1 · Skill practice

IB-style applications · Advanced practice

Quick recap

Related

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Company

Links

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alphy school publishers

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