Exponentials and Logarithms

An exponential function has the variable in the exponent. For this topic, the basic form is

$$y=a^x,$$

where $a$ is a positive constant. The value of the base determines the shape of the graph.

Every graph of $y=a^x$ passes through $(0,1)$ because

$$a^0=1.$$

For every real $x$, the value of $a^x$ is positive. Therefore, the graph never crosses or touches the $x$-axis.

Value of the base	Behaviour of $y=a^x$
$a>1$	The function is increasing. As $x$ increases, $a^x$ grows; as $x$ becomes very negative, $a^x$ approaches $0$.
$a=1$	The function is constant because $1^x=1$, so the graph is the horizontal line $y=1$.
$0<a<1$	The function is decreasing. As $x$ increases, $a^x$ approaches $0$; as $x$ becomes very negative, $a^x$ grows.

For $a>0$ with $a\ne1$, the graph has:

• domain: all real values of $x$;
• range: $y>0$;
• $y$-intercept: $(0,1)$;
• horizontal asymptote: $y=0$.

The graph never reaches its horizontal asymptote. It only gets arbitrarily close to it.

Reciprocal bases produce reflected graphs because

$$\left(\frac{1}{a}\right)^x=a^{-x}.$$

Therefore, the graph of $y=(1/a)^x$ is the reflection of the graph of $y=a^x$ in the $y$-axis.

A useful TMUA observation is that graph behaviour can determine the number of solutions without finding them explicitly. For example, if $a>1$, then $a^x$ is strictly increasing, so an equation $a^x=b$ can have at most one real solution.

Common mistake: a negative base is not part of this graph family. Expressions such as $(-2)^x$ are not real for every real $x$, so they do not define a continuous real-valued exponential function on the whole real line.

A logarithm answers the question: “To what power must the base be raised to obtain this number?”

The statements

$$a^b=c$$

and

$$b=\log_a c$$

mean exactly the same thing.

For example,

$$2^5=32 \quad\Longleftrightarrow\quad \log_2 32=5,$$

$$3^{-3}=\frac{1}{27} \quad\Longleftrightarrow\quad \log_3\left(\frac{1}{27}\right)=-3,$$

and

$$4^{3/2}=8 \quad\Longleftrightarrow\quad \log_4 8=\frac{3}{2}.$$

For real logarithms, the base and argument must satisfy

$$a>0, \qquad a\ne1, \qquad c>0.$$

The base cannot be $1$ because $1^x$ is always $1$, so it cannot produce every positive number. The argument must be positive because $a^x$ is positive for every real $x$ when $a>0$.

Two immediate consequences of the definition are

$$\log_a 1=0$$

because $a^0=1$, and

$$\log_a a=1$$

because $a^1=a$.

The inverse relationship also gives

$$a^{\log_a x}=x \qquad (x>0)$$

and

$$\log_a(a^x)=x.$$

TMUA method: when a logarithm looks unfamiliar, return to index form. For example, instead of trying to recall a special rule for $\log_9 3$, write

$$9^p=3.$$

Since $9=3^2$,

$$3^{2p}=3^1,$$

so $p=1/2$ and therefore

$$\log_9 3=\frac{1}{2}.$$

The logarithm laws follow from the laws of indices. Throughout this section, all logarithms have the same valid base $a$, and every logarithm argument is positive.

The product law is

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

It corresponds to multiplying powers with the same base:

$$a^p\cdot a^q=a^{p+q}.$$

The quotient law is

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

It corresponds to dividing powers with the same base:

$$\frac{a^p}{a^q}=a^{p-q}.$$

The power law is

$$k\log_a x=\log_a(x^k).$$

It corresponds to raising a power to another power:

$$(a^p)^k=a^{kp}.$$

A special case of the quotient and power laws is

$$\log_a\left(\frac{1}{x}\right)=-\log_a x,$$

because

$$\frac{1}{x}=x^{-1}.$$

Worked example

Simplify

$$2\log_3 2+\log_3 9-\log_3 4.$$

Use the power law first:

$$2\log_3 2=\log_3(2^2)=\log_3 4.$$

Then combine the logarithms:

$$\log_3 4+\log_3 9-\log_3 4 =\log_3\left(\frac{4\cdot9}{4}\right) =\log_3 9 =2.$$

The order of operations matters. A coefficient in front of a logarithm becomes a power of its argument; it does not multiply the argument:

$$3\log_a x=\log_a(x^3),$$

not $\log_a(3x)$.

Common traps:

$$\log_a(x+y)\ne\log_a x+\log_a y$$

and

$$\log_a(x-y)\ne\log_a x-\log_a y.$$

There is no logarithm law for splitting a sum or difference inside a logarithm. Also, never combine logarithms with different bases using the product, quotient, or power laws.

Questions requiring the change-of-base formula are not part of this specification, so it is not needed for this topic.

When both sides of an exponential equation can be written with the same base, rewrite them before using logarithms. This is usually the fastest method.

For $a>0$ and $a\ne1$,

$$a^u=a^v \quad\Longrightarrow\quad u=v.$$

This works because $y=a^x$ is one-to-one: each output is produced by only one input.

Worked example

Solve

$$8^{x+1}=4^{2x-1}.$$

Write both sides as powers of $2$:

$$(2^3)^{x+1}=(2^2)^{2x-1}.$$

Therefore,

$$2^{3x+3}=2^{4x-2}.$$

Equate the exponents:

$$3x+3=4x-2,$$

so

$$x=5.$$

Before taking logarithms, look for bases that are powers of a common number. Common rewrites include

$$4=2^2, \qquad 8=2^3, \qquad 9=3^2, \qquad 25=5^2, \qquad 27=3^3.$$

Common mistake: from

$$2^x+2^x=2^{x+1},$$

but not $2^{2x}$. Addition does not follow the multiplication law for indices.

For

$$a^x=b,$$

where $a>0$, $a\ne1$, and $b>0$, the exact solution is

$$x=\log_a b.$$

This is simply the definition of a logarithm.

If $b\le0$, there is no real solution because $a^x$ is always positive. If $b>0$, there is exactly one real solution because $y=a^x$ is strictly increasing when $a>1$ and strictly decreasing when $0<a<1$.

Worked example

Solve

$$3^{2x}=4.$$

Treat the entire exponent $2x$ as the power required to turn $3$ into $4$:

$$2x=\log_3 4.$$

Hence

$$x=\frac{1}{2}\log_3 4.$$

This is an exact answer. No decimal approximation is required.

Equations may need rearranging before they reach the form $a^x=b$.

Worked example

Solve

$$2^{2x+1}=5\cdot2^x.$$

Since $2^x>0$, division by $2^x$ is always valid:

$$2^{x+1}=5.$$

Therefore,

$$x+1=\log_2 5,$$

so

$$x=\log_2 5-1.$$

TMUA method: keep answers in exact logarithmic form unless the question supplies enough information for an exact simplification. On a non-calculator paper, an awkward logarithm is often intended to remain in the answer.

Some exponential equations become ordinary quadratics after a substitution. The key recognition is

$$a^{2x}=(a^x)^2.$$

More generally, powers such as $a^{3x}$, $a^{2x}$, and $a^x$ can be treated as powers of the single quantity $a^x$.

Worked example

Solve

$$25^x-3\cdot5^x+2=0.$$

Since

$$25^x=(5^2)^x=5^{2x}=(5^x)^2,$$

let

$$u=5^x.$$

The equation becomes

$$u^2-3u+2=0.$$

Factorise:

$$(u-1)(u-2)=0.$$

Therefore,

$$u=1 \quad\text{or}\quad u=2.$$

Return to $u=5^x$:

$$5^x=1 \quad\Longrightarrow\quad x=0,$$

and

$$5^x=2 \quad\Longrightarrow\quad x=\log_5 2.$$

Hence the solutions are

$$\boxed{x=0 \quad\text{or}\quad x=\log_5 2}.$$

The substitution has an important restriction:

$$u=a^x>0.$$

Therefore, any zero or negative solution for $u$ must be rejected.

For example, if a substitution produces

$$(u-1)(u+3)=0,$$

then $u=1$ is possible but $u=-3$ is impossible when $u=a^x$.

A reliable method is:

1. Rewrite every exponential term using one base.
2. Identify a repeated quantity such as $a^x$.
3. Substitute $u=a^x$ and record $u>0$.
4. Solve the resulting polynomial equation.
5. Reject non-positive values of $u$.
6. Substitute back and solve each remaining equation of the form $a^x=b$.

Common mistake: solving the quadratic in $u$ is not the end of the problem. The final answers must be values of the original variable $x$.