#1

When you can't trust your math book, come to the pit.

How do you do things like

X^-1/2(x^-3/4)

(x^4=x to the 4th power if that would help anyone)

If anyone thinks they can help me with a few more types of problems i would gladly accept the help.

How do you do things like

X^-1/2(x^-3/4)

(x^4=x to the 4th power if that would help anyone)

If anyone thinks they can help me with a few more types of problems i would gladly accept the help.

#2

O_O

#3

lol...nice answer! :P

#4

I assume that's equal to zero. are you finding x or just simplifying?

Go to the maths/science thread

Go to the maths/science thread

#5

well 3^-2 = 1/9 you do the exponen then make it into a fraction

#6

the one on the bottom is the root, the one on top is the exponent. for instance; x^2/3 would be the cube root of x squared.

#7

X^-1/2 is the reciprocal of the square root of X... The negative means reciprocal of.. Like, 2^-2 is 1/4 because the reciprocal of 4 is 1/4... x^1/2 is the square root... x^1/3 is the cubed root.. Etc etc.. If you get like x^4/3, split it up into (x^1/3)^4 because 1/3 * 4 = 4/3... And that way you can do it without a calculator.. If you have AIM, IM me at CannedBottles for any more questions.

#8

x^1/4 means the 4th root of x

x^1/2 is the square root of x

x^1/2 is the square root of x

#9

no im just simplifying

this teacher also has huge problems that i could easily solve, but she needs it in simpliest form with exponents >.<

I'm getting ready to just find the answer and put it to the first power..

this teacher also has huge problems that i could easily solve, but she needs it in simpliest form with exponents >.<

I'm getting ready to just find the answer and put it to the first power..

#10

X^-1/2(x^-3/4)

I'll get to that one in a second. The negatives add a different aspect.

I'll start with x^1/2. That would be the second root (square root) of x^1.

x^3/4 would then be the fourth root of x^3.

When the exponent is negative, you put it under one.

x^-3/4 would be 1/(fourth root(x^3))

So the answer to that problem would be:

(1/root(x))* (1/(4th root(x^3)))

I'll get to that one in a second. The negatives add a different aspect.

I'll start with x^1/2. That would be the second root (square root) of x^1.

x^3/4 would then be the fourth root of x^3.

When the exponent is negative, you put it under one.

x^-3/4 would be 1/(fourth root(x^3))

So the answer to that problem would be:

(1/root(x))* (1/(4th root(x^3)))

*Last edited by SlashYourFug at Nov 5, 2008,*

#11

(sqrt(x)*4throot(x^3))^-1

Then in 4throot(x^3) Im pretty sure the ^3 cancels out 3 of the roots so:

(sqrt(x)*sqrt(x))^-1= (sqrt(x)^2)^-1 = x^-1

EDIT: just realised they were negative, nevermind me... fixed

Then in 4throot(x^3) Im pretty sure the ^3 cancels out 3 of the roots so:

(sqrt(x)*sqrt(x))^-1= (sqrt(x)^2)^-1 = x^-1

EDIT: just realised they were negative, nevermind me... fixed

*Last edited by EnyoAdonai at Nov 5, 2008,*

#12

isn't that the same as x^-1/2 (times) x^-3/4, in which case it would just be x^-5/4, because multiplying common bases adds exponents?

EDIT: simplified it would be one over the fourth root of x^5, so choice D.

EDIT: simplified it would be one over the fourth root of x^5, so choice D.

#13

I probably should have posted these before but the possible answers are these

A)1overx^3/8

B)x^3/8

C)x^5/4

D)1overx^5/4

I don't see how it gets these answers

I was assuming it was done how one of you where doing it

EDIT: The dude above this got it right but why wouldn't the 4 turn into an 8?

A)1overx^3/8

B)x^3/8

C)x^5/4

D)1overx^5/4

I don't see how it gets these answers

I was assuming it was done how one of you where doing it

EDIT: The dude above this got it right but why wouldn't the 4 turn into an 8?