If m=a^x n=a^yand a²=(m^yn^x)^z Show xyz=1.

The above question is an indices question from an AS textbook. This is question 7 so I will tell you now that this is perhaps a higher difficulty question but you can be assured that Q1-6 will have given you the basics. The key to remember when doing textbooks is that most of the time in progressive questions like this, the solution is a mix of all the knowledge gained in 1-6 for example. They are not isolated questions purposely made impossible to do. It is either related to the previous question or you are supposed to use all the techniques used in Q1-6.

Maths is a subject that is heavily or rather completely dependent on logic. There is ALWAYS a logical method. So the easiest way of cracking questions is finding that method and the rest usually reveals itself.

First you must figure out what type of question you have.

Show xyz=1. There are no numerical values given for x y or z so it is not the type of question where you can just add values into a calculator and show xyz =1.

You will see that the rest of the question gives you more equations.

In questions about proving like this you need to make sure you are starting with or will eventually have equations with xyz in it because that is what you want to prove and want at the end of all your working out.

(TIP: But you can never use the equation xyz =1 to start. This is a common mistake as I did it initially too. There is a rule my maths teacher told me. You never do ‘show that’ questions with the ‘show that’ equation. That would defeat the purpose.)

In this you can see that the equation with xy and z in it, is a²=(m^yn^x)^z

You only want xyz as per question so you want to get rid of all other letters ie a,m and n. The other equations given in the question provides a way for us to get rid of m and n.

Now you might think ‘oh no but I’m still going to have ‘a’’’, but you have to keep in mind that these questions are made in a logical methodical way so that you find the answers in a similarly methodical way, step by step. You can be sure that if you follow the clues, you can solve the case. Therefore you can be sure that since you are going by a sensible logic initially that the rest will eventually unfold itself as you will see.

So you know you can substitute m and n so you do that. You cannot always do somethings all at once, in one go. Since this is maths, there will always be steps and a lot of it as you go higher up the ladder. Moreover you reduce the number of unknowns by doing this, which is also an aim to follow when you are trying to reduce a complicated equation into a simpler one.

You may be more familiar with numbers. But if you can remember and understand the rules then it does not matter if it is number letters or symbols, you just apply the rule.

a²=(a^yxa^yx)^z

Like it always is in maths, from this point here, there is another way you can go from here. One is staying in the bracket, the other is taking the indices out of the bracket. I will do the first method. However you can ask me if you would like to know the other method.

If you look inside the bracket you have one of the rules of indices. You may be more familiar with numbers. But if you can remember and understand the rules then it does not matter if it is number letters or symbols, you just apply the rule.

So if you have the same base number and you are multiplying them (ie a) you can add the indices. So:

a²=(a^yx+yx)^z

a²=(a^2yx)^z

Next remember that you can multiply out indices inside and outside brackets.

a²=a^2yxz

Now if you have the same base on either side of the equation you can equal the indices like so:

2=2xyz

Can you see the end of the tunnel now? J

2/2=xyz

1=xyz

xyz =1

Solve this simultaneous equation: (1) 2x+3y=12 (2) x+4y=11

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If m=a^x, n=a^y, and a^2 =(m^y n^x)^z Show xyz=1.

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