Your goal: Transform expressions with all domain restrictions visible.

Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.

Laws of Exponents, Surds and Logarithms: the key idea

For a nonzero base a and integers r,s, aras=ar+sa^{r} a^{s} = a^{r+s}, (ar)s=ars(a^{r})^{s} = a^{rs}, and a−r=1ara^{-r}=\frac{1}{a^r}. The expression 000^{0} is not assigned a value in these rules. For nonnegative a, a\sqrt{a} is the nonnegative square root, so x2=∣x∣\sqrt{x^{2}} = |x| for real x. A logarithm log⁡bx\log _{b} x asks for the exponent giving x; in real arithmetic b > 0, b ≠ 1 and x > 0. Product and power laws require these domains.

A worked example

Simplify 25x2\sqrt{25x^{2}} for real x.

The square root is nonnegative, so 25x2=5∣x∣\sqrt{25x^{2}} = 5|x|. Writing 5x would fail when x is negative; for x = −2 it would give −10 instead of 10.

Your turn: change one thing

Rationalise 15+2\frac{1}{\sqrt5+2}.

Try this on paper before opening the explanation.

Compare your reasoning

Multiply numerator and denominator by 5−2\sqrt{5} - 2. The denominator becomes 5 − 4 = 1, so the expression equals 5−2\sqrt{5} - 2.

Pause and check

A trap to avoid: State the domain, keep exact values and explain why each step is allowed. A correct answer still needs a reason.

Practise and adjust the level

Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.

Interactive practice loads here. You can also use the complete question set below.

Write a complete argument

For positive a,b, prove log⁡c(ab)=log⁡ca+log⁡cb\log _{c}(ab) = \log _{c} a + \log _{c} b, where c > 0 and c ≠ 1.

Planning hint

List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.

Read the full solution after your attempt

Let u=log⁡cau = \log _{c} a and v=log⁡cbv = \log _{c} b. By definition a=cua = c^{u} and b=cvb = c^{v}. Then ab=cu+vab = c^{u+v}, so the exponent producing ab is u+v. Therefore log⁡c(ab)=u+v\log _{c}(ab) = u+v. Positivity ensures both logarithms are defined.

My proof notebook

Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.

All nine practice questions

Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.

1. Evaluate 2³ × 2⁴.

Foundation

  1. 256
  2. 128
  3. 64
  4. 24
Hint

Add exponents for a common base.

Answer and reasoning

128. 2³×2⁴ = 2⁷ = 128.

2. What is 5−25^{-2}?

Foundation

  1. 1/10
  2. 1/25
  3. −25
  4. −10
Hint

A negative exponent means a reciprocal.

Answer and reasoning

1/25. 5−2=152=1255^{-2}=\frac{1}{5^2}=\frac{1}{25}.

3. For real x, x2\sqrt{x^{2}} equals what?

Core

  1. −x for every real x
  2. x²
  3. |x|
  4. x for every real x
Hint

The principal square root is nonnegative.

Answer and reasoning

|x|. For x ≥ 0 the result is x; for x < 0 it is −x. Together this is |x|.

4. Find log₂ 32.

Core

  1. 5
  2. 4
  3. 6
  4. 16
Hint

Find the exponent of 2 that gives 32.

Answer and reasoning

5. 2⁵ = 32, so log₂ 32 = 5.

5. Which identity is valid for all a,b > 0?

Stretch

  1. log₂(a+b) = log₂ a + log₂ b
  2. a+b=a+b\sqrt{a+b} = \sqrt{a} + \sqrt{b}
  3. (a+b)² = a²+b²
  4. log₂(ab) = log₂ a + log₂ b
Hint

Distinguish product laws from sum claims.

Answer and reasoning

log₂(ab) = log₂ a + log₂ b. Logarithms turn a product of positive inputs into a sum. Taking a=b=1 refutes the logarithmic sum claim; the other two miss cross terms.

6. Simplify (7−2)(7+2)(\sqrt{7} - 2)(\sqrt{7} + 2).

Stretch

  1. 3
  2. 3\sqrt{3}
  3. 11
  4. 7
Hint

Use the difference of squares.

Answer and reasoning

3. The conjugate factors give 7 − 4 = 3.

7. What is 2⁻³?

Foundation

  1. 8
  2. 1/8
  3. −8
  4. −6
Hint

A negative exponent denotes a reciprocal.

Answer and reasoning

1/8. 2⁻³=1/2³=1/8.

8. Simplify 50\sqrt{50} exactly.

Core

  1. 525\sqrt{2}
  2. 25225\sqrt{2}
  3. 10
  4. 252\sqrt{5}
Hint

Factor out the square 25.

Answer and reasoning

525\sqrt{2}. 50=25⋅2=52\sqrt{50}=\sqrt{25\cdot 2}=5\sqrt{2}.

9. For real x, when is x2=x\sqrt{x^{2}}=x?

Stretch

  1. Only when x<0
  2. Never
  3. When x≥0
  4. For every x
Hint

The principal square root is nonnegative.

Answer and reasoning

When x≥0. x2=∣x∣\sqrt{x^{2}}=|x|, which equals x exactly when x≥0.

Choose your next step

Continue to Algebraic identities and factorisation. If this felt difficult, return to a prerequisite above. Every lesson stays open.

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