# Thread: algrebraic negative fractions and indices

1. ## algrebraic negative fractions and indices

Hi,

I am struggling to simplify:

The dot/mark in front of the 4 is meant to be a multiplication sign.

Any help would be great thanks

2. Just note that:

$\displaystyle (a\cdot b)^c = a^c \cdot b^c$ and that $\displaystyle x^a + x^b = x^{a+b}$

You should be able to figure it our easily now.

3. Originally Posted by raphw
Just note that:

$\displaystyle (a\cdot b)^c = a^c \cdot b^c$ and that $\displaystyle x^a + x^b = x^{a+b}$

You should be able to figure it our easily now.
I know that $\displaystyle (a^m)^n = a^m^n$

So on first half of equation I get : $\displaystyle 2x^-^(^1^/^4^) ^\cdot ^(^2^)$

Am I heading in the right direction?

So on first half of equation I get : $\displaystyle 2x^-^(^1^/^4^) ^\cdot ^(^2^)$
Not quite:
[2x^(-1/4)]^2 = 2^(1*2)x^(-1/4 * 2) = 4x^(-1/2)

5. Thanks.

So then $\displaystyle 4x^-^1^/^2 \cdot 4x^3^/^2 = (4x)^1$

Is my thinking correct?

You get a ceegar !!

7. Originally Posted by Wilmer
You get a ceegar !!
Wilmer, can I be a pain?

If X=2

I cannot get the first equation to match 4x ... where am I going wrong?

8. You simply got the answer wrong.

$\displaystyle 4x^-^1^/^2 \cdot 4x^3^/^2 \ne (4x)^1$

but:

$\displaystyle 4x^-^1^/^2 \cdot 4x^3^/^2 = (4)^2(x)^1 = 16x$

9. Thanks Raphw and Wilmer.

Yeah I woke up this morning and realised I missed of one of the fours at the end.

Thanks again

10. Originally Posted by raphw
Just note that:

$\displaystyle (a\cdot b)^c = a^c \cdot b^c$ and that $\displaystyle x^a + x^b = x^{a+b}$
This is incorrect. I presume raphw meant to write $\displaystyle (x^a)(x^b)= x^{a+b}$ with a multiplication on the left side, not a sum.

You should be able to figure it our easily now.