### Overview

This covers; Laws of Indices, Fractional and Negative Powers, Standard Form (building on knowledge gained in Complex Calculations and Accuracy) and Surds. Some of this is quite complicated (some Grade A and A* questions). I think the trick is to make sure you understand the basics thoroughly before moving on to the more advanced stuff.

### Indices and Standard Form Questions

**1.** Work out the value of;

**2. **Work out the value of;

a) (3³ x 5² x 10) x (3^{–2} x 5^{–1} x 10²)

**3. **Calculate these expressions:

a) (10^{–2})^{–2}

b) 9² x ^{3}√27 ÷ ^{4}√3^{20}

**4. **Calculate the value of:-

a) 81^{1/4}

b) 625^{0.25}

**5.** Write the following using index notation:

a) (^{4}√13)^{4}

b) (^{4}√11)^{2}

**6. **Calculate the value of:-

a) 216^{−2÷3}

b) 81^{−3÷4}

**7. **Work out the following with answers in standard form:-

a) (3.2 x 10^{–5}) x (8 x 10^{–3})

b) (3 x 10^{–7}) ÷ (8 x 10^{–1})

c) (1.6 x 10^{4}) + (2.4 x 10^{5})

d) (6.07 x 10^{5}) — (1.8 x 10^{3})

** ****8. **Write this expression in the form x — y√8, x and y are integers:-

(√32 — 2)(√72 + 1)

### Indices and Standard Form Approach

**1.** These look scary but you can just break them down in a few steps. Let’s look at question 1a):-

See final answer to parts a) and b) below

**2. **Again this is worse than it looks. You can just eliminate the brackets and add and subtract the powers (remember that x^{1} = x) so 10 x 10² = 10^{3}

**3. **a) Multiply the indices but remember that a negative times a negative is a positive

b) In this question you need to divide the indices using rule:- ^{x}√W^{y} = W^{(y/x)}

**4. **Start with the knowledge that X^{1/y} = ^{y}√X and for part (b) 0.25 = 1/4, so the same rule can be applied.

**5.** Again, use the rule — ^{x}√W^{y} = W^{(y/x)}

Remember that the question asks for answers in index notation i.e x^{y}

**6. **Yet another rule to use: x^{–y/z} = 1/x^{y/z}

**7. **Work out the following with answers in standard form:-

Remember to give answers in standard form i.e. Y x 10^{n} (with Y a number between 1 and 10 and n is an integer).

**8. **This is most definitely an A* question. I got really confused and it took me some time to crack (more than you get in an exam!) — please look at the detailed steps to see the approach I took.

### Indices and Standard Form Answers

**1.** a)

5^{4} x 3^{3} = 625 x 27 =16,875

b) 6^{4} x 10³ = 1,296,000

**2. **(3³ x 5² x 10) x (3^{–2} x 5^{–1} x 10²) = 3 x 5 x 10^{3} = 15,000

**3. **a) (10^{–2})^{–2} = 10^{4} = 10,000

b) 9² x ^{3}√27 ÷ ^{4}√3^{20}

= 81 x 3 x 3^{5} = 59,049

**4. **a) 81^{1/4} = ^{4}√81 = 3

b) 625^{0.25} = ^{4}√625= 5

**5.** a) (^{4}√13)^{4} = 13^{1} = 13

b) (^{4}√11)^{2} = 11^{2/4} = 11^{0.5}

**6. **a) 216^{−2÷3} = 1/216^{2/3}

= 1/^{3}√216^{2} = 1/6^{2} = 1/36

b) 81^{−3÷4} = 1/81^{3/4}

= 1/^{4}√81^{3} = 1/3^{3} = 1/27

**7. **a) (3.2 x 10^{–5}) x (8 x 10^{–3}) = 25.6 x 10^{–8} =2.56 x 10^{–7}

b) (3 x 10^{–7}) ÷ (8 x 10^{–1}) = 3/8 x 10^{–7}/10^{–1}

=0.375 x 10^{–6} = 3.75 x 10^{–7}

c) (1.6 x 10^{4}) + (2.4 x 10^{5}) = 16,000 + 240,000 = 256,000 = 2.56 x 10^{5}

d) (6.07 x 10^{5}) — (1.8 x 10^{3}) = 607,000 — 1,800 = 605,200 =6.052 x 10^{5}

** ****8. **(√32 — 2)(√72 + 1) = (√4√8 — 2)(√9√8 + 1) = (2√8 — 2)(3√8 + 1)

(When multiplying out the first part of this sum note that: 2√8 x 3√8 = 2 x √8 x √8 x 3 = 6 x 8 = 48)

Therefore (2√8 — 2)(3√8 + 1) = 48 + 2√8 — 6√8 — 2 =

46 — 4√8