Exponentials and Logarithms

What you'll learn

How logarithms “undo” powers, just like square roots undo squares.
How to use log laws to simplify expressions and solve equations.
How to work with $e^x$ and $ln⁡x\ln x$ in growth, decay and cooling models.
How to find ranges, intercepts and tangents for exponential graphs.

1. Powers first: exponential functions

Before logs, you need to be confident with indices. For example, $2^5$ means 2 multiplied by itself 5 times. The number being repeatedly multiplied is called the base.

Definition

Exponential function

An exponential function is a function where the variable appears in the power, such as $y=3^x$ or $y=e^x$ . For AS Maths, the base is positive and not equal to 1.

If both sides of an equation can be written with the same base, you can equate the powers.

Example

Solving when the bases match

Solve $2^{x+1}=64$ .

Rewrite 64 as a power of 2:
$64=2^6$
Equate the powers because the bases are the same:
$x + 1 = 6$
Solve for $x$ :
$x = 5$

2. Logarithms undo exponentials

Most exponential equations do not have matching bases. That is when logarithms are needed.

Definition

Logarithm

$log_a b$ means “the power you put on $a$ to get $b$ ”. So $log_a b=c$ is equivalent to $a^c=b$ . The number $a$ is the base of the logarithm.

Logarithmic form and exponential form contain the same base, power and result arranged differently.

The most common logarithm in AS Pure is the natural logarithm, written $ln⁡x\ln x$ . It means $log_e x$ , where $e$ is the constant approximately 2.718.

The graphs of $y=e^x$ and $y=ln⁡xy=\ln x$ are inverses: they reflect in the line $y = x$ .

$The graphs of y=e^x and y=\ln x are mirror images in the line y=x, showing that the functions are inverses.$

Graphs of y equals e to the x and y equals natural log x showing inverse relationship

Key Idea

Logs reverse powers

If $a^x=b$ , then $x=\log_a b$ . On your calculator, it is usually easiest to use natural logs: $x=ln⁡bln⁡ax=\frac{\ln b}{\ln a}$ .

Example

Solving $5^{x} = 18$

Solve $5^x=18$ , giving your answer to 3 significant figures.

Take natural logs of both sides:
$ln(5^x)=\ln 18$
Use the power law of logs to bring $x$ down:
$xln⁡5=ln⁡18x\ln 5=\ln 18$
Divide by $ln⁡5\ln 5$ :
$x=ln⁡18ln⁡5x=\frac{\ln 18}{\ln 5}$
Evaluate and round:
$x≈1.80x\approx 1.80$

Common Mistake

Dividing inside the log

From $5^x=18$ , do not write $x=ln⁡(18÷5)x=\ln(18\div 5)$ . The correct method is $x=ln⁡18ln⁡5x=\frac{\ln 18}{\ln 5}$ .

3. The log laws

Log laws let you combine or split logarithms. They are especially useful when solving equations.

The three log laws match multiplication, division and powers inside a logarithm to addition, subtraction and multiplication outside.

Definition

Log laws

For $M > 0$ , $N > 0$ , $a > 0$ and $a≠1a\neq 1$ :

log⁡a(MN)=log⁡aM+log⁡aNlog⁡a(MN)=log⁡aM−log⁡aNlog⁡a(Mp)=plog⁡aM\begin{aligned} \log_a(MN)&=\log_a M+\log_a N\\ \log_a\left(\frac{M}{N}\right)&=\log_a M-\log_a N\\ \log_a(M^p)&=p\log_a M \end{aligned}

Example

Writing as a single logarithm

Express $2\log_a(x+3)-\log_a 7$ as a single logarithm.

Use the power law on the first term:
$2log⁡a(x+3)=log⁡a((x+3)2)2\log_a(x+3)=\log_a\left((x+3)^2\right)$
Use the subtraction law:
$log⁡a((x+3)2)−log⁡a7=log⁡a((x+3)27)\log_a\left((x+3)^2\right)-\log_a 7=\log_a\left(\frac{(x+3)^2}{7}\right)$
State the restriction from the log input:
$x > - 3$

Example

Solving a log equation

Solve $2\log_2 x-\log_2 5=3$ .

Use the power law:
$2\log_2 x=\log_2(x^2)$
Combine into one logarithm:
$log⁡2(x25)=3\log_2\left(\frac{x^2}{5}\right)=3$
Convert from log form to exponential form:
$x25=23\frac{x^2}{5}=2^3$
Solve, remembering $x > 0$ because $log_2 x$ is only defined for positive $x$ :
$x=210x=2\sqrt{10}$

Common Mistake

Logs do not split over addition

$log_a(M+N)$ is not the same as $log_a M+\log_a N$ . Log laws work with multiplication, division and powers, not addition inside the log.

Sometimes you are told $y=\log_2 x$ and asked to rewrite related expressions.

Example

Using $y = lo g_{2} x$

Given $y=\log_2 x$ , write these in terms of $y$ : $log_2(x^2)$ , $log_2(8x)$ and $log_4 x$ .

Use the power law:
$log_2(x^2)=2\log_2 x=2y$
Write 8 as $2^3$ , then use the multiplication law:
$log_2(8x)=\log_2 8+\log_2 x=3+y$
For base 4, remember $4=2^2$ :
$log⁡4x=log⁡2xlog⁡24=y2\log_4 x=\frac{\log_2 x}{\log_2 4}=\frac{y}{2}$

4. Solving equations involving $ln⁡x\ln x$

The argument of a logarithm is the expression inside it. For $ln⁡(3x−2)\ln(3x-2)$ , the argument is $3 x - 2$ . The argument must be positive.

$For \ln(3x-2) the input must satisfy 3x-2>0, so the logarithm is only defined to the right of x=\frac23.$

Common Mistake

Log inputs must be positive

You can only take $ln⁡\ln$ or $log⁡\log$ of a positive number. Always check the argument, especially when a solution has come from squaring or from a quadratic.

Example

Solving a natural log equation

Solve $ln⁡(3x−2)=1\ln(3x-2)=1$ .

Exponentiate both sides to undo $ln⁡\ln$ :
$3x-2=e^1$
Rearrange:
$x=e+23x=\frac{e+2}{3}$
Check the argument is positive:
$3 x - 2 = e > 0$

You can also use logs to solve simultaneous equations by substituting one expression into the other.

Example

A simultaneous equation with $e$ and $ln$

Solve to 2 decimal places:

e2y=x+2,ln⁡(x−1)=2y−1e^{2y}=x+2,\qquad \ln(x-1)=2y-1

Take logs of the first equation:
$2y=ln⁡(x+2)2y=\ln(x+2)$
Substitute into the second equation:
$ln⁡(x−1)=ln⁡(x+2)−1\ln(x-1)=\ln(x+2)-1$
Combine the logarithms:
$ln⁡(x−1x+2)=−1\ln\left(\frac{x-1}{x+2}\right)=-1$
Exponentiate and solve for $x$ :
$x−1x+2=e−1\frac{x-1}{x+2}=e^{-1}$
Rearranging gives:
$x=e+2e−1≈2.75x=\frac{e+2}{e-1}\approx 2.75$
Substitute into $2y=ln⁡(x+2)2y=\ln(x+2)$ :
$y≈0.78y\approx 0.78$

5. Equations that become quadratics

Some equations contain both $e^y$ and $e^{-y}$ . Since $e−y=1eye^{-y}=\frac{1}{e^y}$ , a substitution often turns the equation into a quadratic.

$The substitution u=e^y changes terms involving e^y and e^{-y} into algebraic terms in u.$

Example

Using the substitution $u = e^{y}$

Solve exactly:

3e^y+10e^{-y}=11

Let $u=e^y$ , so $e−y=1ue^{-y}=\frac{1}{u}$ and $u > 0$ .
Substitute and multiply by $u$ :
$3u+10u=113u+\frac{10}{u}=11$
Rearrange into a quadratic:
$3u^2-11u+10=0$
Factorise:
$(3 u - 5) (u - 2) = 0$
Convert back to $y$ :
$y=ln⁡(53)ory=ln⁡2y=\ln\left(\frac{5}{3}\right)\quad \text{or}\quad y=\ln 2$

6. Exponential models and rates of change

Exponential models often look like $P=Ae^{kt}$ . Here, $A$ is the initial amount and $k$ controls the rate of growth or decay.

$In P=Ae^{kt}, the sign of k determines whether the model grows or decays from the initial value A.$

If $k > 0$ , the model grows. If the exponent is negative, such as $Ae^{-kt}$ , the model decays.

Key Idea

Derivative of an exponential model

If $P=Ae^{kt}$ , then $dPdt=kAekt\frac{dP}{dt}=kAe^{kt}$ . The rate of change is proportional to the current amount.

Example

A growth model

A plant population is modelled by $P=60e^{0.07t}$ , where $t$ is measured in weeks.

Find the initial population by putting $t = 0$ :
$P(0)=60e^0=60$
Differentiate to find the rate of increase:
$dPdt=60⋅0.07e0.07t=4.2e0.07t\frac{dP}{dt}=60\cdot 0.07e^{0.07t}=4.2e^{0.07t}$
At 10 weeks, the rate is:
$4.2e0.7≈8.464.2e^{0.7}\approx 8.46$
To find when the population exceeds 300, solve the inequality:
$60e^{0.07t}>300$
Take logs and solve:
$t>ln⁡50.07≈22.99t>\frac{\ln 5}{0.07}\approx 22.99$

Tip

Interpreting model answers

If time is continuous, “22.99 weeks” is fine to 2 decimal places. If the question asks for a whole number of weeks, round up to 23 weeks because the population must exceed the target.

For cooling or radioactive decay, you are often given one data point to find the constant $k$ .

Example

Finding a decay constant

A temperature model is $T=70e^{-kt}+18$ . After 4 minutes, $T = 55$ . Find $k$ and then find when $T = 40$ .

The cooling model decreases towards the ambient temperature asymptote T=18, with the given and target temperatures marked.

Substitute $t = 4$ and $T = 55$ :
$55=70e^{-4k}+18$
Rearrange:
$e−4k=3770e^{-4k}=\frac{37}{70}$
Take logs:
$k=−14ln⁡(3770)≈0.159k=-\frac{1}{4}\ln\left(\frac{37}{70}\right)\approx 0.159$
Now set $T = 40$ :
$40=70e^{-kt}+18$
Solve for $t$ :
$t=−ln⁡(1135)k≈7.27t=-\frac{\ln\left(\frac{11}{35}\right)}{k}\approx 7.27$

7. Exponential graphs, range, intercepts and tangents

The graph of $y=e^{x+2}-4$ is the graph of $y=e^x$ shifted left 2 and down 4. It has a horizontal asymptote at $y = - 4$ .

The transformed exponential graph has horizontal asymptote y=-4, intercepts, and a tangent at the y-intercept.

Graph of y equals e to the x plus 2 minus 4 showing asymptote, intercepts and tangent

Definition

Range and asymptote

The range is the set of possible output values of a function. A horizontal asymptote is a horizontal line that the curve approaches but never reaches.

Example

Range, intercepts and tangent

Let $f(x)=e^{x+2}-4$ .

Since $e^{x+2}>0$ , the function is always greater than -4:
$f (x) > - 4$
The range is:
$y > - 4$
Find the y-intercept by putting $x = 0$ :
$A=(0,e^2-4)$
Find the x-intercept by putting $f (x) = 0$ :
$e^{x+2}=4$
Take logs:
$B=(ln⁡4−2,0)B=(\ln 4-2,0)$
Differentiate:
$f'(x)=e^{x+2}$
At $A$ , the gradient is $e^2$ , so the tangent is:
$y-(e^2-4)=e^2x$

Exam technique

In the exam

Check whether bases match before reaching for logs.
When solving log equations, combine logs into a single logarithm before converting to exponential form.
In modelling questions, identify the initial value, differentiate for rate questions, and use logs to solve for time.

Self review

Check yourself

Can you explain why $e^x$ is always positive?
Can you solve an equation containing both $e^y$ and $e^{-y}$ using a substitution?
Can you state the range of $y=e^{3x}-5$ without sketching the graph?

Exponentials and Logarithms

What you'll learn

How logarithms “undo” powers, just like square roots undo squares.
How to use log laws to simplify expressions and solve equations.
How to work with $e^x$ and $ln⁡x\ln x$ in growth, decay and cooling models.
How to find ranges, intercepts and tangents for exponential graphs.

1. Powers first: exponential functions

Before logs, you need to be confident with indices. For example, $2^5$ means 2 multiplied by itself 5 times. The number being repeatedly multiplied is called the base.

Definition

Exponential function

An exponential function is a function where the variable appears in the power, such as $y=3^x$ or $y=e^x$ . For AS Maths, the base is positive and not equal to 1.

If both sides of an equation can be written with the same base, you can equate the powers.

Example

Solving when the bases match

Solve $2^{x+1}=64$ .

Rewrite 64 as a power of 2:
$64=2^6$
Equate the powers because the bases are the same:
$x + 1 = 6$
Solve for $x$ :
$x = 5$

2. Logarithms undo exponentials

Most exponential equations do not have matching bases. That is when logarithms are needed.

Definition

Logarithm

$log_a b$ means “the power you put on $a$ to get $b$ ”. So $log_a b=c$ is equivalent to $a^c=b$ . The number $a$ is the base of the logarithm.

Logarithmic form and exponential form contain the same base, power and result arranged differently.

The most common logarithm in AS Pure is the natural logarithm, written $ln⁡x\ln x$ . It means $log_e x$ , where $e$ is the constant approximately 2.718.

The graphs of $y=e^x$ and $y=ln⁡xy=\ln x$ are inverses: they reflect in the line $y = x$ .

$The graphs of y=e^x and y=\ln x are mirror images in the line y=x, showing that the functions are inverses.$

Graphs of y equals e to the x and y equals natural log x showing inverse relationship

Key Idea

Logs reverse powers

If $a^x=b$ , then $x=\log_a b$ . On your calculator, it is usually easiest to use natural logs: $x=ln⁡bln⁡ax=\frac{\ln b}{\ln a}$ .

Example

Solving $5^{x} = 18$

Solve $5^x=18$ , giving your answer to 3 significant figures.

Take natural logs of both sides:
$ln(5^x)=\ln 18$
Use the power law of logs to bring $x$ down:
$xln⁡5=ln⁡18x\ln 5=\ln 18$
Divide by $ln⁡5\ln 5$ :
$x=ln⁡18ln⁡5x=\frac{\ln 18}{\ln 5}$
Evaluate and round:
$x≈1.80x\approx 1.80$

Common Mistake

Dividing inside the log

From $5^x=18$ , do not write $x=ln⁡(18÷5)x=\ln(18\div 5)$ . The correct method is $x=ln⁡18ln⁡5x=\frac{\ln 18}{\ln 5}$ .

3. The log laws

Log laws let you combine or split logarithms. They are especially useful when solving equations.

The three log laws match multiplication, division and powers inside a logarithm to addition, subtraction and multiplication outside.

Definition

Log laws

For $M > 0$ , $N > 0$ , $a > 0$ and $a≠1a\neq 1$ :

log⁡a(MN)=log⁡aM+log⁡aNlog⁡a(MN)=log⁡aM−log⁡aNlog⁡a(Mp)=plog⁡aM\begin{aligned} \log_a(MN)&=\log_a M+\log_a N\\ \log_a\left(\frac{M}{N}\right)&=\log_a M-\log_a N\\ \log_a(M^p)&=p\log_a M \end{aligned}

Example

Writing as a single logarithm

Express $2\log_a(x+3)-\log_a 7$ as a single logarithm.

Use the power law on the first term:
$2log⁡a(x+3)=log⁡a((x+3)2)2\log_a(x+3)=\log_a\left((x+3)^2\right)$
Use the subtraction law:
$log⁡a((x+3)2)−log⁡a7=log⁡a((x+3)27)\log_a\left((x+3)^2\right)-\log_a 7=\log_a\left(\frac{(x+3)^2}{7}\right)$
State the restriction from the log input:
$x > - 3$

Example

Solving a log equation

Solve $2\log_2 x-\log_2 5=3$ .

Use the power law:
$2\log_2 x=\log_2(x^2)$
Combine into one logarithm:
$log⁡2(x25)=3\log_2\left(\frac{x^2}{5}\right)=3$
Convert from log form to exponential form:
$x25=23\frac{x^2}{5}=2^3$
Solve, remembering $x > 0$ because $log_2 x$ is only defined for positive $x$ :
$x=210x=2\sqrt{10}$

Common Mistake

Logs do not split over addition

$log_a(M+N)$ is not the same as $log_a M+\log_a N$ . Log laws work with multiplication, division and powers, not addition inside the log.

Sometimes you are told $y=\log_2 x$ and asked to rewrite related expressions.

Example

Using $y = lo g_{2} x$

Given $y=\log_2 x$ , write these in terms of $y$ : $log_2(x^2)$ , $log_2(8x)$ and $log_4 x$ .

Use the power law:
$log_2(x^2)=2\log_2 x=2y$
Write 8 as $2^3$ , then use the multiplication law:
$log_2(8x)=\log_2 8+\log_2 x=3+y$
For base 4, remember $4=2^2$ :
$log⁡4x=log⁡2xlog⁡24=y2\log_4 x=\frac{\log_2 x}{\log_2 4}=\frac{y}{2}$

4. Solving equations involving $ln⁡x\ln x$

The argument of a logarithm is the expression inside it. For $ln⁡(3x−2)\ln(3x-2)$ , the argument is $3 x - 2$ . The argument must be positive.

$For \ln(3x-2) the input must satisfy 3x-2>0, so the logarithm is only defined to the right of x=\frac23.$

Common Mistake

Log inputs must be positive

You can only take $ln⁡\ln$ or $log⁡\log$ of a positive number. Always check the argument, especially when a solution has come from squaring or from a quadratic.

Example

Solving a natural log equation

Solve $ln⁡(3x−2)=1\ln(3x-2)=1$ .

Exponentiate both sides to undo $ln⁡\ln$ :
$3x-2=e^1$
Rearrange:
$x=e+23x=\frac{e+2}{3}$
Check the argument is positive:
$3 x - 2 = e > 0$

You can also use logs to solve simultaneous equations by substituting one expression into the other.

Example

A simultaneous equation with $e$ and $ln$

Solve to 2 decimal places:

e2y=x+2,ln⁡(x−1)=2y−1e^{2y}=x+2,\qquad \ln(x-1)=2y-1

Take logs of the first equation:
$2y=ln⁡(x+2)2y=\ln(x+2)$
Substitute into the second equation:
$ln⁡(x−1)=ln⁡(x+2)−1\ln(x-1)=\ln(x+2)-1$
Combine the logarithms:
$ln⁡(x−1x+2)=−1\ln\left(\frac{x-1}{x+2}\right)=-1$
Exponentiate and solve for $x$ :
$x−1x+2=e−1\frac{x-1}{x+2}=e^{-1}$
Rearranging gives:
$x=e+2e−1≈2.75x=\frac{e+2}{e-1}\approx 2.75$
Substitute into $2y=ln⁡(x+2)2y=\ln(x+2)$ :
$y≈0.78y\approx 0.78$

5. Equations that become quadratics

Some equations contain both $e^y$ and $e^{-y}$ . Since $e−y=1eye^{-y}=\frac{1}{e^y}$ , a substitution often turns the equation into a quadratic.

$The substitution u=e^y changes terms involving e^y and e^{-y} into algebraic terms in u.$

Example

Using the substitution $u = e^{y}$

Solve exactly:

3e^y+10e^{-y}=11

Let $u=e^y$ , so $e−y=1ue^{-y}=\frac{1}{u}$ and $u > 0$ .
Substitute and multiply by $u$ :
$3u+10u=113u+\frac{10}{u}=11$
Rearrange into a quadratic:
$3u^2-11u+10=0$
Factorise:
$(3 u - 5) (u - 2) = 0$
Convert back to $y$ :
$y=ln⁡(53)ory=ln⁡2y=\ln\left(\frac{5}{3}\right)\quad \text{or}\quad y=\ln 2$

6. Exponential models and rates of change

Exponential models often look like $P=Ae^{kt}$ . Here, $A$ is the initial amount and $k$ controls the rate of growth or decay.

$In P=Ae^{kt}, the sign of k determines whether the model grows or decays from the initial value A.$

If $k > 0$ , the model grows. If the exponent is negative, such as $Ae^{-kt}$ , the model decays.

Key Idea

Derivative of an exponential model

If $P=Ae^{kt}$ , then $dPdt=kAekt\frac{dP}{dt}=kAe^{kt}$ . The rate of change is proportional to the current amount.

Example

A growth model

A plant population is modelled by $P=60e^{0.07t}$ , where $t$ is measured in weeks.

Find the initial population by putting $t = 0$ :
$P(0)=60e^0=60$
Differentiate to find the rate of increase:
$dPdt=60⋅0.07e0.07t=4.2e0.07t\frac{dP}{dt}=60\cdot 0.07e^{0.07t}=4.2e^{0.07t}$
At 10 weeks, the rate is:
$4.2e0.7≈8.464.2e^{0.7}\approx 8.46$
To find when the population exceeds 300, solve the inequality:
$60e^{0.07t}>300$
Take logs and solve:
$t>ln⁡50.07≈22.99t>\frac{\ln 5}{0.07}\approx 22.99$

Tip

Interpreting model answers

If time is continuous, “22.99 weeks” is fine to 2 decimal places. If the question asks for a whole number of weeks, round up to 23 weeks because the population must exceed the target.

For cooling or radioactive decay, you are often given one data point to find the constant $k$ .

Example

Finding a decay constant

A temperature model is $T=70e^{-kt}+18$ . After 4 minutes, $T = 55$ . Find $k$ and then find when $T = 40$ .

The cooling model decreases towards the ambient temperature asymptote T=18, with the given and target temperatures marked.

Substitute $t = 4$ and $T = 55$ :
$55=70e^{-4k}+18$
Rearrange:
$e−4k=3770e^{-4k}=\frac{37}{70}$
Take logs:
$k=−14ln⁡(3770)≈0.159k=-\frac{1}{4}\ln\left(\frac{37}{70}\right)\approx 0.159$
Now set $T = 40$ :
$40=70e^{-kt}+18$
Solve for $t$ :
$t=−ln⁡(1135)k≈7.27t=-\frac{\ln\left(\frac{11}{35}\right)}{k}\approx 7.27$

7. Exponential graphs, range, intercepts and tangents

The graph of $y=e^{x+2}-4$ is the graph of $y=e^x$ shifted left 2 and down 4. It has a horizontal asymptote at $y = - 4$ .

The transformed exponential graph has horizontal asymptote y=-4, intercepts, and a tangent at the y-intercept.

Graph of y equals e to the x plus 2 minus 4 showing asymptote, intercepts and tangent

Definition

Range and asymptote

The range is the set of possible output values of a function. A horizontal asymptote is a horizontal line that the curve approaches but never reaches.

Example

Range, intercepts and tangent

Let $f(x)=e^{x+2}-4$ .

Since $e^{x+2}>0$ , the function is always greater than -4:
$f (x) > - 4$
The range is:
$y > - 4$
Find the y-intercept by putting $x = 0$ :
$A=(0,e^2-4)$
Find the x-intercept by putting $f (x) = 0$ :
$e^{x+2}=4$
Take logs:
$B=(ln⁡4−2,0)B=(\ln 4-2,0)$
Differentiate:
$f'(x)=e^{x+2}$
At $A$ , the gradient is $e^2$ , so the tangent is:
$y-(e^2-4)=e^2x$

Exam technique

In the exam

Check whether bases match before reaching for logs.
When solving log equations, combine logs into a single logarithm before converting to exponential form.
In modelling questions, identify the initial value, differentiate for rate questions, and use logs to solve for time.

Self review

Check yourself

Can you explain why $e^x$ is always positive?
Can you solve an equation containing both $e^y$ and $e^{-y}$ using a substitution?
Can you state the range of $y=e^{3x}-5$ without sketching the graph?

Exponentials and Logarithms

What you'll learn

1. Powers first: exponential functions

2. Logarithms undo exponentials

3. The log laws

4. Solving equations involving $ln⁡x\ln x$

5. Equations that become quadratics

6. Exponential models and rates of change

7. Exponential graphs, range, intercepts and tangents

Recap questions

You've reached the end

Exponentials and Logarithms

What you'll learn

1. Powers first: exponential functions

2. Logarithms undo exponentials

3. The log laws

4. Solving equations involving $ln⁡x\ln x$

5. Equations that become quadratics

6. Exponential models and rates of change

7. Exponential graphs, range, intercepts and tangents

Recap questions

You've reached the end

Exponentials and Logarithms

What you'll learn

1. Powers first: exponential functions

2. Logarithms undo exponentials

3. The log laws

4. Solving equations involving ln⁡x\ln xlnx

5. Equations that become quadratics

6. Exponential models and rates of change

7. Exponential graphs, range, intercepts and tangents

Recap questions

You've reached the end

Exponentials and Logarithms Revision Guide

Exponentials and Logarithms

What you'll learn

1. Powers first: exponential functions

2. Logarithms undo exponentials

3. The log laws

4. Solving equations involving ln⁡x\ln xlnx

5. Equations that become quadratics

6. Exponential models and rates of change

7. Exponential graphs, range, intercepts and tangents

Recap questions

You've reached the end

Exponentials and Logarithms Revision Guide

4. Solving equations involving $ln⁡x\ln x$

4. Solving equations involving $ln⁡x\ln x$