Indices

RosaA

4

These questions are confusing. :s-smilie:

I basically don't know how to work out the answer... :redface:

Help me please someone!!! :colondollar:

x = 3^m and y = 3ⁿ

Express in terms of x and y:
a) 3^m+n
b) 3²ⁿ

Thank you! :biggrin:

Reply 1

Zacken

22

Original post by RosaA

These questions are confusing. :s-smilie:

I basically don't know how to work out the answer... :redface:

Help me please someone!!! :colondollar:

x = 3^m and y = 3ⁿ

Express in terms of x and y:
a) 3^m+n
b) 3²ⁿ

Thank you! :biggrin:

Okay, so you need to use this rules.

a^x \times a^y = a^{x+y}

. That is, if the 'bases' are the same and the powers different, when you multiply the two numbers, you add the powers.

Conversely, if you're adding the powers you can split

a^{x+y}

into

a^x a^y

. In this case, you're working with

a=3

- can you take it from here?

For the second part, you also need to make use of the fact that

(a^x)^y = a^{xy}

. So:

(2^2)^3 = 2^{2\times 3} = 2^{6}

for example. You can verify this yourself with a calculator.

In this case, we have

3^{2n} = (3^n)^2

.

Reply 2

Zacken

22

Original post by RosaA

...

Here's a quick intuitive feel for

a^x \times a^y = a^{x+y}

. I'll do it with

a=3

here to make life simpler, but it works for any

a

:

\displaystyle 3^x = \underbrace{3 \times 3 \times \cdots \times 3}_{x \, \text{times}}

and

3^y = \underbrace{3 \times 3 \times \cdots \times 3}_{y \, \text{times}}

So:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} 3^x \times 3^y = \underbrace{3 \times 3 \times \cdots \times 3}_{x \, \text{times}} \times \underbrace{3 \times 3 \times \cdots \times 3}_{y \, \text{times}} = \underbrace{3 \times 3 \times \cdots \times 3}_{x+y \, \text{times}}\end{equation*}

You now have a product of

x

3

's and

y

3

's and there are

x+y

threes being multiplied in total.

Hence

Unparseable latex formula:

3^x \times 3^y = \underbrace{3 \times 3 \times \cdots \times 3}_{x+y\,\text{times}}\end{equation*} = 3^{x+y}

as expected.

(edited 8 years ago)

Reply 3

Zacken

22

Original post by RosaA

...

And here's a quick intuitive argument as to why

(a^x)^y

is

a^{xy}

, it'll use the other theorem we 'proved' above.

You're comfortable that

a^x \times a^y = a^{x+y}

now, I hope.

So, since:

\displaystyle (a^x)^y = \underbrace{a^x \times a^x \times \cdots \times a^x}_{y \, \text{times}}

then we can simply add their powers:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} (a^x)^y = a^{\overbrace{x + x + \cdots + x}^{y \, \text{times}}} = a^{xy}\end{equation*}

Since

\underbrace{x + x + \cdots + x}_{y \, \text{times}}

(i.e: x added to itself y times) is just

xy

.

Reply 4

RosaA

OP

4

Original post by Zacken

Okay, so you need to use this rules.

a^x \times a^y = a^{x+y}

. That is, if the 'bases' are the same and the powers different, when you multiply the two numbers, you add the powers.

Conversely, if you're adding the powers you can split

a^{x+y}

into

a^x a^y

. In this case, you're working with

a=3

- can you take it from here?

For the second part, you also need to make use of the fact that

(a^x)^y = a^{xy}

. So:

(2^2)^3 = 2^{2\times 3} = 2^{6}

for example. You can verify this yourself with a calculator.

In this case, we have

3^{2n} = (3^n)^2

.

Yup, that makes more sense now. :smile:

I didn't realise that if the "bases" are the same you could simply just add the powers like normal. I just need 2 get remember that now and I'll be fine. Thank you!

In terms of the second question, I understand it now.
I've just realised how simple it was *laughs*

I didn't occur to me to use any of the index laws :colondollar:

-which was the issue, so thank you for going over them for me :biggrin:

Reply 5

Zacken

22

Original post by RosaA

Yup, that makes more sense now. :smile:

I didn't realise that if the "bases" are the same you could simply just add the powers like normal. I just need 2 get remember that now and I'll be fine. Thank you!

In terms of the second question, I understand it now.
I've just realised how simple it was *laughs*

I didn't occur to me to use any of the index laws :colondollar:

-which was the issue, so thank you for going over them for me :biggrin:

Haha, no problem - I wrote all of that in seperate posts so I can link back to this thread when other people have the same questions instead of me having to type it up all over again, hope you didn't mind! :smile:

Glad that helped!! :h:

Reply 6

RosaA

OP

4

Original post by Zacken

Haha, no problem - I wrote all of that in seperate posts so I can link back to this thread when other people have the same questions instead of me having to type it up all over again, hope you didn't mind! :smile:

Glad that helped!! :h:

That's fine by me - :smile:

Reply 7

Zacken

22

Original post by RosaA

That's fine by me - :smile:

Awesome.

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