This note covers powers, roots and standard form: the index laws, negative and fractional powers, and how to write and calculate with very large and very small numbers. These ideas appear in algebra, geometry, science and almost every calculator question, so fluency here saves time everywhere else.
Key ideas
A power tells you how many times a number is multiplied by itself. In \(a^n\), \(a\) is the base and \(n\) is the index (or power). A root undoes a power: \(\sqrt{49}=7\) because \(7^2=49\), and \(\sqrt[3]{64}=4\) because \(4^3=64\).
Learn the squares up to \(15^2\), the cubes \(1^3,2^3,3^3,4^3,5^3\) and \(10^3\), and the powers of 2 up to \(2^{10}\). Knowing these makes simplifying much faster.
The index laws work for any non-zero base \(a\) and any rational indices:
- Multiplying powers of the same base: add the indices.
- Dividing powers of the same base: subtract the indices.
- A power of a power: multiply the indices.
- Anything (non-zero) to the power 0 is 1.
- A negative index means a reciprocal. It does not make the answer negative.
- A fractional index means a root. The denominator of the fraction is the root, and the numerator is the power.
Standard form writes a number as \(a\times10^{k}\) with \(1\le a\lt10\) and \(k\) an integer. Large numbers have positive \(k\). Numbers between 0 and 1 have negative \(k\). For example \(450\,000=4.5\times10^{5}\) and \(0.00032=3.2\times10^{-4}\).
To multiply or divide in standard form, deal with the numbers and the powers of ten separately. To add or subtract, first rewrite both numbers with the same power of ten. Always check that your final answer has \(1\le a\lt10\); if not, adjust the power of ten.
On a calculator, use the key marked EXP or \(\times10^{x}\) to enter standard form. Do not type \(\times\), \(10\) and the power separately unless you use brackets, because the order of operations can give a wrong answer.
Index equations. When the unknown is in the index, write both sides as powers of the same base and then equate the indices. For example, to solve \(2^{x}=\dfrac{1}{32}\), write \(\dfrac{1}{32}=2^{-5}\). Then \(2^{x}=2^{-5}\), so \(x=-5\). To solve \(3^{2x-1}=81\), write \(81=3^{4}\). Then \(2x-1=4\), so \(x=2.5\).
Comparing sizes. To compare two numbers in standard form, compare the powers of ten first. The number with the larger power is larger. Only if the powers are equal do you compare the mantissas. For example, \(9.1\times10^{3}\) is smaller than \(1.2\times10^{4}\), even though 9.1 is bigger than 1.2.
Roots and indices together. Square roots and cube roots can be written as indices, so \(\sqrt{x}=x^{1/2}\) and \(\sqrt[3]{x^{2}}=x^{2/3}\). Writing roots as indices lets you use the index laws. For example, \(\sqrt{x}\times x^{3/2}=x^{1/2+3/2}=x^{2}\). Remember also that every positive number has two square roots, one positive and one negative, but the symbol \(\sqrt{\ }\) means the positive one.
Powers of ten and prefixes. Standard form works well with units. For example, 1 km is \(10^{3}\) m and 1 mm is \(10^{-3}\) m. If a question gives lengths in different units, convert them to the same unit before multiplying or dividing, and then use the index laws on the powers of ten.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(a^m\times a^n=a^{m+n}\) | Same base, multiplying: add indices. |
| \(a^m\div a^n=a^{m-n}\) | Same base, dividing: subtract indices. |
| \((a^m)^n=a^{mn}\) | Power of a power: multiply indices. |
| \(a^0=1\) | True for every non-zero \(a\). |
| \(a^{-n}=\dfrac{1}{a^n}\) | Negative index: take the reciprocal. |
| \(a^{1/n}=\sqrt[n]{a}\) | Fractional index of the form \(1/n\): the \(n\)th root. |
| \(a^{m/n}=\left(\sqrt[n]{a}\right)^m\) | Take the root first, then the power. This keeps numbers small. |
| \((ab)^n=a^nb^n\) | The power applies to every factor inside the bracket, including numbers. |
| \(a\times10^{k}\), \(1\le a\lt10\) | Standard form. \(k\) is a positive or negative integer. |
Worked examples
Example 1 Work out (a) \(125^{2/3}\) and (b) \(\left(\dfrac{4}{9}\right)^{-3/2}\), giving each answer as an integer or a fraction.
(a) The denominator 3 means cube root, the numerator 2 means square. Do the root first: \(\sqrt[3]{125}=5\).
Then square: \(5^2=25\).
(b) The negative index means turn the fraction upside down: \(\left(\dfrac{4}{9}\right)^{-3/2}=\left(\dfrac{9}{4}\right)^{3/2}\).
The denominator 2 means square root: \(\sqrt{\dfrac{9}{4}}=\dfrac{3}{2}\).
The numerator 3 means cube: \(\left(\dfrac{3}{2}\right)^3=\dfrac{27}{8}\).
Answer: (a) \(25\); (b) \(\dfrac{27}{8}\).
Example 2 Simplify \(\dfrac{(2a^3b)^4}{8a^5b^2}\).
Expand the bracket in the numerator. The power 4 applies to the 2, to \(a^3\) and to \(b\): \((2a^3b)^4=2^4\,a^{12}\,b^4=16a^{12}b^4\).
Now divide numbers: \(16\div8=2\).
Subtract the indices of \(a\): \(12-5=7\). Subtract the indices of \(b\): \(4-2=2\).
Answer: \(2a^7b^2\).
Example 3 The Sun is \(1.5\times10^{8}\) km from the Earth. Light travels at \(3.0\times10^{5}\) km/s. (a) Find the time light takes to reach the Earth, in standard form. (b) Find the sum \(6.2\times10^{4}+3.5\times10^{5}\) in standard form.
(a) Time = distance \(\div\) speed \(=\dfrac{1.5\times10^{8}}{3.0\times10^{5}}\).
Divide the numbers: \(1.5\div3.0=0.5\). Subtract the powers: \(8-5=3\). So the time is \(0.5\times10^{3}\) s.
This is not yet in standard form because \(0.5\lt1\). Write \(0.5\times10^{3}=5\times10^{2}\).
(b) Use the same power of ten. Write \(6.2\times10^{4}=0.62\times10^{5}\).
Add: \(0.62\times10^{5}+3.5\times10^{5}=4.12\times10^{5}\).
Answer: (a) \(5\times10^{2}\) seconds (about 8.33 minutes); (b) \(4.12\times10^{5}\).
Common errors
- Thinking a negative index gives a negative number. \(2^{-3}=\dfrac18\), not \(-8\).
- Adding indices when the bases are different, such as writing \(2^3\times3^2=6^5\). The index laws only work for the same base.
- Forgetting that a power outside a bracket applies to the number too: \((3x^2)^3=27x^6\), not \(3x^6\).
- Leaving an answer such as \(12\times10^{4}\) or \(0.4\times10^{6}\). Neither is in standard form.
- Adding standard form numbers by adding the mantissas and the powers separately.
Exam tip: for fractional indices, take the root before the power, as in \(8^{2/3}=(\sqrt[3]{8})^2=4\). Show the intermediate step so that a slip on the calculator does not hide your method.
Check yourself
- Write \(0.000\,072\) in standard form.
Answer
Move the point 5 places right to get 7.2, so the power is negative: \(7.2\times10^{-5}\). - Evaluate \(8^{-2/3}\).
Answer
\(8^{-2/3}=\dfrac{1}{8^{2/3}}=\dfrac{1}{(\sqrt[3]{8})^2}=\dfrac{1}{2^2}=\dfrac14\). - Work out \((3.6\times10^{5})\times(2.5\times10^{-3})\) and give the answer in standard form.
Answer
\(3.6\times2.5=9\) and \(10^{5}\times10^{-3}=10^{2}\). So the product is \(9\times10^{2}\) (which is 900). - Simplify \(\dfrac{(3x^2y)^2}{6x^3y^{-1}}\).
Answer
Numerator: \(9x^4y^2\). Divide: \(9\div6=\dfrac32\). Subtract indices: \(x^{4-3}=x\) and \(y^{2-(-1)}=y^3\). Answer: \(\dfrac{3}{2}xy^3\).