A Level Maths: Index Laws
Introduction
You would have met index laws or the rules of indices as they are more commonly known at GCSE.
They can be summarised as follows:
1.x^m \times x^n=x^{m+n}
For this rule to work it is important that we have the same base.
For instance, here we have a base of x:
x^2 \times x^5=x^{2+5}=x^7However:
x^5 \times y^3=x^5 \times y^3In the above instance it is not possible to add the powers together as the bases are not the same.
2. x^m \div x^n=x^{m-n}
Just for the first rule, this rule also requires that the bases are the same.
x^{12} \div x^5=x^{12-5}=x^7But:
x^{12} \div y^5=x^{12} \div y^5More Laws
3. \left(x^m\right)^n=x^{m \times n}=x^{m n}
For the third rule we have a base to a power raised to another power.
So:
\left(x^4\right)^5=x^{4 \times 5}=x^{20}What is important here is to be careful when you have a situation such as the following:
\left(2 x^4\right)^5The biggest mistake that is made here is to not raise the 2 to the power 5. It is important to remember that everything inside the bracket is being raised to the power 5.
In this instance therefore we will obtain the following as the answer:
\begin{aligned} & 2^5=32 \\ & \left(x^4\right)^5=x^{20} \end{aligned}And so the final answer is 32 x^{20}
4. a^0=1
This is a rule that you need to remember is that anything to the power zero is always equal to. If you think about this for a moment it seems a little strange but we will go through the reason as to why.
Consider:
\frac{x^5}{x^3}=x^{5-3}=x^2From the above we have applied the subtraction rule. We also know that dividing a number by itself is equal to 1. Such as:
\frac{2^1}{2^1}=1Applying the subtraction rule on the left we have:
2^0=1And so anything to the power zero is always equal to 1.
Exam Style Questions
In order to answer this question successfully you will need to apply the third rule as detailed above and to remember that everything inside the bracket needs to be raised to the power on the outside of the bracket.
You also need to be able to show your working out and not be over dependent on your calculator.
16^{\frac{3}{4}}=\left(16^{\frac{1}{4}}\right)^3=2^3=8This is something that you can quickly determine with your calculator but in A Level Maths it is important that you are showing all of your working.
\left(a^{12}\right)^{\frac{3}{4}}=a^{12 \times \frac{3}{4}}=a^9The overall final answer is therefore 8 a^9
Let us first tackle the denominator:
\left(4 x^5 y\right)^3=64 x^{15} y^3Now for the denominator:
\left(2 x y^2\right) \times\left(8 x^{10} y^4\right)=16 x^{11} y^6We now have the following where we can apply the division rule:
\frac{64 x^{15} y^3}{16 x^{11} y^6}=4 x^4 y^{-3}=\frac{4 x^4}{y^3}How you write the “y term” does not matter unless of course a question is asking for the answer in a particular format.
Negative And Fractional Index Laws
\sqrt{a}=a^{\frac{1}{2}}You need to remember what a square root is as a fractional power.
\sqrt{9}=9^{\frac{1}{2}}=3
\sqrt[n]{x}=x^{\frac{1}{n}}This is for the nth root of a value and knowing how to write this as a fractional power is very important.
\sqrt[5]{32}=32^{\frac{1}{5}}=2\sqrt[m]{x^n}=x^{\frac{n}{m}}\sqrt[2]{8^3}=8^{\frac{3}{2}}=\left(2^3\right)^{\frac{3}{2}}=2^{\frac{9}{2}}
You sometimes need to express a term in terms of a certain base and your calculator will not be able to do this for you. With regards to the last example the value a calculator will give you is 22.627417. It does not give you an answer in terms of base 2 as shown above.
a^{-n}=\frac{1}{a^n}4^{-2}=\frac{1}{4^2}=\frac{1}{16}
Being able to write negative powers in terms of a more simple expression is again an important skill to have especially if you need to simplify an expression or to obtain an answer in a certain format.
Exam Questions
The instructions in the question are quite clear, the answer needs to be in the form of base 2 but if you put this directly into your calculator you get 20.1587368! So how do you get an answer in terms of base 2 only?
The solution is all about breaking down the numbers as much as possible.
Clearly nothing can be done with 5 or 3. But what about the 4 and the 16?
\begin{aligned} &4^{\frac{2}{3}}=\left(2^2\right)^{\frac{2}{3}}=2^{\frac{4}{3}}\\ &16^{\frac{1}{3}}=\left(2^4\right)^{\frac{1}{3}}=2^{\frac{4}{3}} \end{aligned}This now gives us the following where we now have a common term which can now be factorised.
\begin{aligned} & 5 \times 2^{\frac{4}{3}}+3 \times 2^{\frac{4}{3}} \\ & 2^{\frac{4}{3}}(5+3) \\ & 2^{\frac{4}{3}}(8) \\ & 2^{\frac{4}{3}}\left(2^3\right) \\ & 2^{\frac{13}{3}} \end{aligned}The key with this question is to have first recognised what numbers can be written as base 2 and then to use the appropriate rules of indices to obtain them in terms of the correct fractional power. It is important to then recognise that a common term exists and factorisation can occur. What is inside the bracket can be written as base 2 and then finally the addition rule of indices can be applied.
It is these types of questions that are covered during our upcoming A Level Maths Revision Course Christmas 2023 in London. And it is these types of questions that cannot be done on a calculator. You will be taught the various techniques needed to succeed at A Level.
If you are looking for more regular support then you may also wish to seek the regular guidance of an online maths tutor. Lessons can be arranged once a week or as often as required.
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