• Apr 25th 2011, 03:50 AM
dumluck
I'm fairly clear on the rules govening the multiplication of exponents with the same base, but what about adding them?

I.E. 2^5 + 2^5 + 3^6 + 3^6 + 3^6 = ?

The first part 2^5 + 2^5 is 2^6 because one is adding one 2^5 to another 2^5

but using this logic would 3^5 + 3^5 + 3^5 equal? 3^7?

The latter is incorrect, it should be 3^6? but why?
• Apr 25th 2011, 04:02 AM
Prove It
It's because you need to see repeated addition as multiplication.

2^5 + 2^5 is the same as "2 lots of 2^5".

2(2^5) = (2^1)(2^5) = 2^(1 + 5) = 2^6.

For the second, you have 3^5 + 3^5 + 3^5, in other words "3 lots of 3^5".

3(3^5) = (3^1)(3^5) = 3^(1 + 5) = 3^6.
• Apr 25th 2011, 04:02 AM
pickslides
The base is 2 in the first, adding 2 numbers and 3 in the second adding 3 numbers.
• Apr 25th 2011, 04:17 AM
dumluck
Thank Prove it..

So does that mean if we have 2^5 + 2^5 + 2^5 + 2^5 , it would be 2^8 as there are four lots of 2?

2.2.2(2^5) = 2^2+^6 = 2^8?
• Apr 25th 2011, 04:18 AM
Prove It
Correct :)
• Apr 25th 2011, 04:35 AM
Soroban
Hello, dumluck!

Quote:

2^5 + 2^5 + 3^6 + 3^6 + 3^6 = ?

The first part 2^5 + 2^5 is 2^6 because one is adding one 2^5 to another 2^5

but using this logic would 3^5 + 3^5 + 3^5 equal 3^7 ?

The latter is incorrect, it should be 3^6? but why?

Here's a baby-talk approach which should clear away any fog . . .

How do we add x + x ?

Factor: .x + x .= .x(1 + 1) .= .x(2) .= .2x

Then: .3^5 + 3^5 + 3^5 .= .3^5(1 + 1 + 1) .= .(3^5)(3) .= .3^6

And: .2^5 + 2^5 + 2^5 + 2^5 .= .2^5(1 + 1 + 1 + 1)

. . . . = .(2^5)(4) .= .(2^5)(2^2) .= .2^7

• Apr 25th 2011, 04:49 AM
dumluck
Quote:

Originally Posted by Soroban
Hello, dumluck!

Here's a baby-talk approach which should clear away any fog . . .

How do we add x + x ?

Factor: .x + x .= .x(1 + 1) .= .x(2) .= .2x

Then: .3^5 + 3^5 + 3^5 .= .3^5(1 + 1 + 1) .= .(3^5)(3) .= .3^6

And: .2^5 + 2^5 + 2^5 + 2^5 .= .2^5(1 + 1 + 1 + 1)

. . . . = .(2^5)(4) .= .(2^5)(2^2) .= .2^7

So I was wrong above ; 2^5 + 2^5 + 2^5 + 2^5 = (2^5)(4) = (2^5)(2)(2) = (2^5)(2^2) ; when you multiply exponents you add the exponent so it equals 2^7

So 3^6 + 3^6 + 3^6 + 3^6 = 4(3^6) ????

I would assume therefore that you cannot add exponents with different bases, the same way you cannot multiply them?
• Apr 25th 2011, 06:05 AM
dumluck
[QUOTE=dumluck;643221]So 3^6 + 3^6 + 3^6 + 3^6 = 4(3^6) ????
QUOTE]

ok, i'll at least try this!!... 3^2.3^1.3^6 , is it therefore 3^9?
• Apr 25th 2011, 06:18 AM
topsquark
[QUOTE=dumluck;643247]
Quote:

Originally Posted by dumluck
So 3^6 + 3^6 + 3^6 + 3^6 = 4(3^6) ????
QUOTE]

ok, i'll at least try this!!... 3^2.3^1.3^6 , is it therefore 3^9?

No, you were right the first time. (4)(3^6) = (2^2)(3^6). We cannot simplify this, in general.

Note that if the bases are equal we can apply (a^n)(a^m) = a^(n + m)
If the exponents are equal we can apply (a^n)(b^n) = (ab)^n

But we have neither situation in this case.

-Dan
• Apr 25th 2011, 06:32 AM
dumluck
[QUOTE=topsquark;643252]
Quote:

Originally Posted by dumluck
No, you were right the first time. (4)(3^6) = (2^2)(3^6). We cannot simplify this, in general.

Note that if the bases are equal we can apply (a^n)(a^m) = a^(n + m)
If the exponents are equal we can apply (a^n)(b^n) = (ab)^n

But we have neither situation in this case.

-Dan

Hi Dan,

So for example (9)(3^4) would be ok as it can be broken into 3^2.3^1? so 9 additions of 3^4 would be 3^7?
• Apr 25th 2011, 08:17 AM
topsquark
[QUOTE=dumluck;643256]
Quote:

Originally Posted by topsquark

Hi Dan,

So for example (9)(3^4) would be ok as it can be broken into 3^2.3^1? so 9 additions of 3^4 would be 3^7?

Close.
http://latex.codecogs.com/png.latex?...^{2 + 4} = 3^6

That's the second time you've included an extra factor. I'm not sure why you are doing it, but at least you are doing it consistently. :) Notice that 9 is not equal to 3^2 * 3^1 = 27. I'm not sure why you are including the 3^1 in there.

-Dan
• Apr 25th 2011, 08:42 AM
dumluck
[QUOTE=topsquark;643309]
Quote:

Originally Posted by dumluck
Close.
http://latex.codecogs.com/png.latex?...^{2 + 4} = 3^6

That's the second time you've included an extra factor. I'm not sure why you are doing it, but at least you are doing it consistently. :) Notice that 9 is not equal to 3^2 * 3^1 = 27. I'm not sure why you are including the 3^1 in there.

-Dan

fatigue :). I think I'm ok. (9)(3^6) = (3^6 + 3^6 + 3^6 + 3^6 + 3^6 + 3^6 + 3^6 + 3^6 + 3^6) = (3^2)(3^6) = (3^8) is that right?
• Apr 25th 2011, 09:05 AM
topsquark
[QUOTE=dumluck;643327]
Quote:

Originally Posted by topsquark

fatigue :). I think I'm ok. (9)(3^6) = (3^6 + 3^6 + 3^6 + 3^6 + 3^6 + 3^6 + 3^6 + 3^6 + 3^6) = (3^2)(3^6) = (3^8) is that right?

Yup. (Nod)

-Dan