Transformations of Exp/Log (AA SL)

Shifting, stretching, or reflecting an exponential or log graph follows the same rules as any other function, but the asymptote makes the effect easy to check: track where it lands and the rest of the curve follows. This page focuses on describing transformations and tracking the asymptote through them. It's part of the broader Exponentials & Logarithms topic.

10 questions on this sub-topic.

Practise exp/log transformations → Try exam-style questions

Exponent laws you'll lean on

Covered under IB syllabus reference SL1.5, which introduces logarithms with base 10 and \(e\), and the equivalence \(a^x=b \iff \log_a b = x\). Transformation questions often need these two exponent rules to simplify the transformed equation first.

Product rule

\(a^m \cdot a^n = a^{m+n}\)

Not in the formula booklet. Useful for rewriting a horizontal shift, like \(2^{x-3}\), as a vertical stretch: \(2^{x-3}=2^x\cdot2^{-3}\).

Power rule

\((a^m)^n = a^{mn}\)

Not in the formula booklet. Multiplies exponents rather than stacking them - relevant whenever a transformed exponential is raised to a further power.

Need the full syllabus wording and the log-base introduction? See Exponentials & Logarithms.

Worked examples

1
Medium
No calc
[4 marks]

The graph of \(y=e^{x}\) is translated to give \(y=e^{x-2}+1\).

(a) Describe the transformations.
(b) State the equation of the horizontal asymptote.

Worked solution

(a) Comparing \(e^{x-2}+1\) with \(e^x\): \(x\to x-2\) shifts right 2. A1

The \(+1\) shifts up 1. A1

(b) The original asymptote \(y=0\) shifts up 1: \(y\). R1

\(=1\). A1

A1 Right 2 A1 Up 1 R1 Asymptote shifts A1 \(y=1\)
2
Medium
No calc
[2 marks]

Describe the transformation taking \(y=2^{x}\) to \(y=2^{x-3}-1\).

Worked solution

Comparing \(2^{x-3}-1\) with \(2^x\): replacing \(x\) by \(x-3\) shifts right 3. A1

The \(-1\) shifts down 1. A1

Translation by \(\begin{pmatrix}3\\-1\end{pmatrix}\).

A1 Right 3 A1 Down 1
3
Hard
No calc
[6 marks]

\(f(x)=x^2.\) Find the equation of \(y=f(x)\) after reflection in the \(x\)-axis, vertical stretch factor 3, then translation up 5; give it in expanded form.

Worked solution

\(y = -x^2.\) M1 A1
\(y = -3x^2.\) M1 A1
\(y = -3x^2 + 5.\) M1 A1

M1 Reflection A1 \(-x^2\) M1 Stretch A1 \(-3x^2\) M1 Translate A1 \(-3x^2+5\)

Common mistakes

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Quick answers

How do you find the equation of a transformed exponential graph?

Apply each transformation in the order given to the base function: a reflection changes the sign in front, and a translation replaces \(x\) with \((x - h)\) and adds the vertical shift \(k\) outside.

What happens to the asymptote when you transform an exponential graph?

A vertical translation moves the horizontal asymptote by the same amount; a horizontal translation leaves it unchanged, since the asymptote is a \(y\)-value, not an \(x\)-value.

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