F1, F3 - Exponential and logarithmic functions | AQA A-Level Mathematics

Exponential functions turn an input into a power; logarithmic functions recover that input. Use F1 to recognise exponential graphs and F3 to find their inverses, including the domains, ranges and transformations of the natural functions.

Positive-base exponentials

If $4^x=1/8$ , can the exponent be both negative and fractional, and can a graph explain why there is only one answer?

Start with repeated multiplication: the outputs of $2^x$ at $x=0,1,2,3$ are $1,2,4,8$ . Each step of one in the input multiplies the output by $2$ . Moving backwards divides by $2$ , so $2^{-1}=1/2$ , while fractional inputs fill the gaps: $2^{1/2}=\sqrt2$ .

You already know

For $a>0$ , $a^0=1$ , $a^{-r}=1/a^r$ and $a^{1/2}=\sqrt a$ . These index results are not in the formula booklet: learn them.

An exponential function has the variable in the exponent. The function $f(x)=a^x$ , with $a>0$ , is defined for every real $x$ . For $a>1$ , it increases; for $0<a<1$ , it decreases. When $a\ne1$ , its range is $(0,\infty)$ , its intercept is $(0,1)$ , and its horizontal asymptote is $y=0$ : the curve approaches this line without reaching it. These graph facts are not supplied in the formula booklet.

The boundary cases matter. If $a=1$ , the function is constant with range $\{1\}$ , so it has no inverse on $\mathbb R$ . A negative base does not define a real-valued function on all real inputs: $(-2)^{1/2}$ is not real.

Increasing and decreasing exponential curves and their inverse logarithms

The left panel shows the two directions of change. The right panel will let us reverse those same input-output pairs.

Worked example 1: a decreasing exponential

Question. For $f(x)=(1/3)^x$ , find the values at $x=-1,0,1$ and state the features needed for a sketch.

Plan. Evaluate three powers, then use the base to decide the direction and limiting behaviour.

\begin{aligned} f(-1)&=3 && \text{take the reciprocal} \\ f(0)&=1 && \text{zero power} \\ f(1)&=1/3 && \text{first power} \end{aligned}

Why this step? A base between $0$ and $1$ multiplies each new output by a fraction, so the curve decreases while remaining positive.

Answer. Sketch a smooth decreasing curve through $(-1,3)$ , $(0,1)$ and $(1,1/3)$ , approaching $y=0$ to the right. Domain: $\mathbb R$ ; range: $(0,\infty)$ .

Check. Moving from $-1$ to $0$ multiplies $3$ by $1/3$ , giving $1$ ; the next step gives $1/3$ .

Now change just the base to its reciprocal. The intercept and range stay the same, but the direction reverses; in fact $(1/3)^{-x}=3^x$ , a reflection in the $y$ -axis.

Worked example 2: reversing the base

Question. For $g(x)=3^x$ , find the values at $x=-1,0,1$ and state its sketch features.

Plan. Use the same inputs and compare the base with $1$ .

\begin{aligned} g(-1)&=1/3 && \text{negative power} \\ g(0)&=1 && \text{zero power} \\ g(1)&=3 && \text{first power} \end{aligned}

Why this step? The base $3>1$ gives an increasing function, so its approach to zero is now on the left.

Answer. The curve increases through $(-1,1/3)$ , $(0,1)$ and $(1,3)$ , with asymptote $y=0$ , domain $\mathbb R$ and range $(0,\infty)$ .

Check. Each output is three times the previous one. The outputs match example 1 when the input signs are reversed.

A negative exponent changes the size of a positive output, not its sign. This is especially useful when checking whether a sketch belongs above the axis.

Watch out

Wrong: $3^{-1}=-3$ . A negative exponent means a reciprocal, so $3^{-1}=1/3>0$ . An unshifted positive-base exponential never crosses the $x$ -axis.

Change the base again in your own calculation. State the full range, rather than only the values you evaluated.

Inverse logarithms

A logarithm tells you which exponent produces a given positive output.

Since $4^2=16$ , the statement $\log_4 16=2$ records the same input-output pair in reverse. The argument $16$ is positive, but the logarithm itself need not be positive.

Logarithm

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

\log_a v=u\quad\Longleftrightarrow\quad a^u=v

The base is $a$ ; the argument is $v$ . For example, $\log_2(1/4)=-2$ . There is no real $\log_2 0$ or $\log_2(-4)$ , because a positive-base power is strictly positive.

The exponential is one-to-one when $a\ne1$ , so each positive output gives exactly one exponent. Its inverse graph is reflected in $y=x$ : $\log_a x$ has domain $(0,\infty)$ , range $\mathbb R$ , intercept $(1,0)$ and vertical asymptote $x=0$ . It increases when $a>1$ and decreases when $0<a<1$ . The definition, inverse identities and graph facts here are not supplied in the formula booklet.

Worked example 3: a fractional exponent

Question. Find the exact value of $\log_4 8$ .

Plan. Name the exponent and express both sides using base $2$ .

\begin{aligned} 4^u&=8 && \text{inverse definition} \\ 2^{2u}&=2^3 && \text{common base} \\ 2u&=3 && \text{one-to-one exponential} \\ u&=3/2 && \text{divide by }2 \end{aligned}

Why this step? Equal powers of the same valid base have equal exponents; guessing integer powers would miss a fractional answer.

Answer. $\log_4 8=3/2$ .

Check. $4^{3/2}=(\sqrt4)^3=2^3=8$ .

Change only the argument to its reciprocal. The corresponding exponent must change sign, because taking a reciprocal negates a power.

Worked example 4: answering the opening puzzle

Question. Find the exact value of $\log_4(1/8)$ .

Plan. Convert to an exponential equation and again compare powers of $2$ .

\begin{aligned} 4^u&=1/8 && \text{inverse definition} \\ 2^{2u}&=2^{-3} && \text{reciprocal power} \\ 2u&=-3 && \text{equal exponents} \\ u&=-3/2 && \text{divide by }2 \end{aligned}

Why this step? The positive argument is below $1$ , so an increasing exponential reaches it at a negative input.

Answer. $\log_4(1/8)=-3/2$ , the unique solution of the opening equation.

Check. $4^{-3/2}=1/8$ , so the answer is negative and fractional without giving a negative output.

The inverse identities are $\log_a(a^u)=u$ for real $u$ , and $a^{\log_a v}=v$ for $v>0$ . Their different restrictions come from the domain of the inner logarithm.

Watch out

Wrong: $\log_4(-8)=-3/2$ . The exponent $-3/2$ gives $1/8$ , not $-8$ . A negative logarithm is allowed; a negative real logarithm argument is not.

Use the reciprocal idea once more, with a new argument.

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F1, F3 - Exponential and logarithmic functions

Positive-base exponentials

Inverse logarithms

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