Candidly speaking, since I started working on indices I haven't seen nor come accros a problem that could produce different bases as your own. As you made 9^(n+2) *3^(n+2) out of 27^(n+2)

Now I think the method that will be applicable to a problem is the one that must be used.

Thank very much Bobbym, God bless you.

bobbym

2013-08-12 04:05:41

All that is correct.

Why didn't you use 3^(3n) * 3^(6) but rather wrote 9^(n+2) * 3^(n+2).

Because it was easier to cancel.

EbenezerSon

2013-08-12 03:56:21

Now I have learnt that! Thanks much Bobbym!

Okay, back to my question.

27^(n+2) =27^n * 27^2 = 3^(3n) * 3^(6) .

Why didn't you use 3^(3n) * 3^(6) but rather wrote 9^(n+2) * 3^(n+2).

Thanks in advance.

bobbym

2013-08-12 03:34:05

All of that is correct but I have to interpret every bit of it.

When you write 27^n+3 in mathematics that means

It is properly written 27^(n+3) this means

Notice they are both very different.

EbenezerSon

2013-08-12 03:30:07

27^n+2=27^n * 27^2

ThÄ±s is how I mean, I will learn parenthesis in it proper way as you say.

For instance, 6^n+3=6^n * 6^3.

Because, for instance, 3^2 * 3^2=3^(2+2).

What do you say.

bobbym

2013-08-12 03:13:25

You should learn to latex or to use parentheses better.

That modern notation they are using in that book is not good.

EbenezerSon

2013-08-12 03:10:29

In fact I have not seen an indicial problem being split to get different numbers as the base. like what is in #269.

.ThiÅŸ problem is from indices.

bobbym

2013-08-12 03:02:00

I am sorry, I can not follow that. Please bracket it off.

EbenezerSon

2013-08-12 03:00:55

I suppose the base must always be the same in each case. So I percieved it to be,