Skip to content
A Level Maths: Harder Indices Questions
Harder Indices – Introduction
In this article we are going to look at some more questions involving harder indices but they are more challenging than what you may have seen. They also cannot be done with a calculator. Not that they cannot be solved with a calculator, they can, but a calculator will give the answer in the desired form.
This is quite often the case when doing A Level Maths, that you cannot be too over reliant on your calculator and so you need to be able to understand the necessary processes that are involved in all areas of mathematics at A Level.
Harder Indices - Example 1
Solution
In order to simplify this question it is important to be able to write each each of the above terms in the smallest base possible such as base 2 or base 3 or even a combination of both for instance.
\begin{aligned} & 6^{\frac{5}{2}} \times 12^{-\frac{1}{2}} \times 18^{-\frac{3}{2}} \\ & 6^{\frac{5}{2}}=(2 \times 3)^{\frac{5}{2}}=2^{\frac{5}{2}} \times 3^{\frac{5}{2}} \\ & 12^{-\frac{1}{2}}=(4 \times 3)^{-\frac{1}{2}}=4^{-\frac{1}{2}} \times 3^{-\frac{1}{2}}=\frac{1}{2} \times 3^{-\frac{1}{2}}=2^{-1} \times 3^{-\frac{1}{2}} \\ & 18^{-\frac{3}{2}}=(9 \times 2)^{-\frac{3}{2}}=9^{-\frac{3}{2}} \times 2^{-\frac{3}{2}}=\left(9^{\frac{1}{2}}\right)^{-3} \times 2^{-\frac{3}{2}}=3^{-3} \times 2^{-\frac{3}{2}} \end{aligned}You can see from above that everything is in terms of base 2 and base 3. Writing everything in terms of these bases will now give:
2^{\frac{5}{2}} \times 3^{\frac{5}{2}} \times 2^{-1} \times 3^{-\frac{1}{2}} \times 3^{-3} \times 2^{-\frac{3}{2}}Bringing these together will give us:
\begin{aligned} & =\left(2^{\frac{5}{2}} \times 2^{-1} \times 2^{-\frac{3}{2}}\right) \times\left(3^{\frac{5}{2}} \times 3^{-\frac{1}{2}} \times 3^{-3}\right) \\ & =2^0 \times 3^{-1}=3^{-1}=\frac{1}{3} \end{aligned}Harder Indices - Example 2
We are asked to simplify the above and again this would need to be done without a calculator. If you were to put the above into your calculator then you obtain the answer 7.071067812
Want we want here is a common term. So let us consider the terms individually and the rule of indices that we will be using is: a^m \times a^n=a^{m+n}
The first terms can be written as 2^{\frac{5}{2}}=2^{\frac{1}{2}} \times 2^2 and the second will stay as 2^{\frac{1}{2}}
You will see that with the first term we have broken down the power. In other words, we have used the rule of indices that you would have met at GCSE, backwards.
You can see that 2^{\frac{1}{2}} is the common term. Now that we have a common term we can now factorise as follows: 2^{\frac{1}{2}} \times 2^2+2^{\frac{1}{2}}=2^{\frac{1}{2}}\left(2^2+1\right)=5 \sqrt{2}
Harder Indices – Question Practice
Try the following questions on your own and look at the solutions to compare your answers.
Answers
First we have the following:
\begin{aligned} & 3^{\frac{1}{2}} \times 3^2-3^{\frac{1}{2}} \\ & 3^{\frac{1}{2}}\left(3^2-1\right) \\ & =8 \sqrt{3} \end{aligned}Second we have the following solution: \begin{aligned} & 2^{\frac{1}{2}}+\left(2^{\frac{1}{2}} \times 2\right)+\left(2^{\frac{1}{2}} \times 2^2\right) \\ & 2^{\frac{1}{2}}\left(1+2+2^2\right) \\ & =7 \sqrt{2} \end{aligned}
Finally we have the following solution: \frac{\sqrt{y}}{1}-\frac{1}{\sqrt{y}}
Which can be written as a single algebraic fraction as follows: \frac{y-1}{\sqrt{y}}
Harder Indices Example
Here is another slightly awkward question involving indices:
Again you need to make sure that you are fully aware of all the rules of indices that you learnt when doing GCSE Maths. For this particular question the rule of indices that will be used is:
\left(a^m\right)^n=a^{m n}The main way to start is to ask who to rewrite the number 9?
\left(3^2\right)^{3 x+2}Now just multiply the powers together: 3^{6 x+4} \text { ie } 3^y \text { where } y=6 x+4
Harder Indices Example
This question looks awful to look at, at first sight. Your approach to this question is again to use the rules of indices.
\begin{aligned} & a^{m+n}=a^m \times a^n \\ & a^{m n}=\left(a^m\right)^n \end{aligned}Consider 2^{2 x+1}. The question that needs to be asked is how else can it be written?
\begin{aligned} & 2^{2 x} \times 2 \\ & \left(2^x\right)^2 \times 2 \end{aligned}Now you will see that 2^x is common, so we have 2\left(2^x\right)^2-5\left(2^x\right)-12=0. And what we have here is a quadratic in disguise and this can be solved using a simple substitution such as y=2^x to give:
\begin{aligned} & 2 y^2-5 y-12=0 \\ & (2 y+3)(y-4)=0 \\ & y=-\frac{3}{2} \quad y=4 \\ & 2^x=-\frac{3}{2} \quad 2^x=4 \quad \rightarrow x=2 \\ & x \log 2=\log \left(-\frac{3}{2}\right) \end{aligned}This means that x=2
is the only solution because you cannot take a negative log as this will result in a maths error.
You have seen some more challenging questions and you need to remember that when you do your A Level Maths exam you are potentially going to see questions like the above. What you need to remember is that these questions are simply asking you to use the rules of indices that you learnt at GCSE but backwards. Doing this is a key skill at A Level and the best way to guarantee success is to be trying as many of these questions as possible.
Try all the questions here again without looking at the solution and see if you agree. As well as using the rules of indices you are also looking at breaking down terms into their smallest base.