Intermediate Algebra: simplify x ^ 1/2 / x ^ 3/4 * x ^ -4
- Thread starter Helen
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Re: Intermediate Algebra
Is this your question:
\(\displaystyle \frac{x^{1/2}}{x^{3/4}\cdot x^{-4}}\)
Just apply your basic exponent rules:
\(\displaystyle a^{m} \cdot a^{n} = a^{m+n}\)
\(\displaystyle \frac{a^{m}}{a^{n}} = a^{m-n}\)
These ones are all you need for this particular question. Simplify the denominator first using the above rules.
Is this your question:
\(\displaystyle \frac{x^{1/2}}{x^{3/4}\cdot x^{-4}}\)
Just apply your basic exponent rules:
\(\displaystyle a^{m} \cdot a^{n} = a^{m+n}\)
\(\displaystyle \frac{a^{m}}{a^{n}} = a^{m-n}\)
These ones are all you need for this particular question. Simplify the denominator first using the above rules.
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- Feb 4, 2004
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Did the tutor guess your meaning correctly? (Note: The lack of grouping symbols was again the source of the confusion.)Helen said:, Thank you for your help.
Since there are no grouping nor exponent symbols, I'm afraid your meaning is again unclear. For instance, is x multiplied by the fraction 15/4? Or is 15/4 a power? Or is just 15 a power, and 4 divides the x-term? Or does the 4 divide the fraction? And so forth.Helen said:
Kindly please re-read the formatting articles mentioned in the "Read Before Posting" thread for this forum, and then reply with clarification. Thank you!
Eliz.
Eliz.
did have the question right.
This is the way that I worked it.
x ^ 1/2 / x ^ 3/4 * x - 4 = x ^ 1/2 / x - (- 1 ) = x ^ 1/2 - ( - 1) = 2 / 2 = 1
3 * - 4 = - 12 + 3 = 1 / 15 / 4 ( - 4 + - 1 = - 3 )
I hope that this can be understood.
Helen
did have the question right.This is the way that I worked it.
x ^ 1/2 / x ^ 3/4 * x - 4 = x ^ 1/2 / x - (- 1 ) = x ^ 1/2 - ( - 1) = 2 / 2 = 1
3 * - 4 = - 12 + 3 = 1 / 15 / 4 ( - 4 + - 1 = - 3 )
I hope that this can be understood.
Helen
- Joined
- Feb 4, 2004
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- 16,582
Working from the formatting in the previous tutor's post, I will guess that the above means the following:Helen said:
x ^ 1/2 / x ^ 3/4 * x - 4 = x ^ 1/2 / x - (- 1 ) = x ^ 1/2 - ( - 1) = 2 / 2 = 1
. . . . .[ x^(1/2) ] / [ x^(3/4) x^(-4) ]
. . . . .[ x^(1/2) ] / [x^(-(-1)) ]
. . . . .[ x^(1/2) ]^(-(-1))
. . . . .2/2
. . . . .1
However, I'm afraid I don't understand the reasoning behind any of the above...? And I'm afraid I can't figure out what the various arithmetic computations in the following line might be for. I think they're related to some aspect of the exponents, but I'm not sure...? Sorry!
You might want to review your exponent rules:
. . . . .a) x[sup:27aepl78]m[/sup:27aepl78] x[sup:27aepl78]n[/sup:27aepl78] = x[sup:27aepl78]m+n[/sup:27aepl78]
. . . . .b) x[sup:27aepl78]m[/sup:27aepl78] / x[sup:27aepl78]n[/sup:27aepl78] = x[sup:27aepl78]m-n[/sup:27aepl78]
. . . . .c) (x[sup:27aepl78]m[/sup:27aepl78])[sup:27aepl78]n[/sup:27aepl78] = x[sup:27aepl78]mn[/sup:27aepl78]
Using these, how should you simplify the following?
. . . . .x[sup:27aepl78]3/4[/sup:27aepl78] x[sup:27aepl78]-4[/sup:27aepl78]
Once you have the simplification of the denominator (using rule (a) above), simplify the entire fractional expression (using rule (b) above).
Note: Your final answer should be in the form "x^(some power)". You should not be getting a numerical value for this! :shock:
Eliz.
Helen, if you want to get PROPER help, then YOU need to post PROPERly;Helen said:
your expression PROPERly typed should be:
x^(1/2) / [x^(3/4) * x^(-4)]
There is a big difference(as example) with 40 / 2 * 4 and 40 / (2 * 4):
40 / 2 * 4 = 80 and 40 / (2 * 4) = 5 : UNDERSTAND?
You were told that x^m * x^n = x^(m + n); so:
[x^(3/4) * x^(-4)] = x^(3/4 + (-4)) = x^(-13/4)
So now you're left with x^(1/2) / x^(-13/4) = x^(1/2 - (-13/4)) = x^(1/2 + 13/4) = x^(15/4)
You did at one point SHOW 15/4, but your typed steps were impossible to follow.
If you keep on typing stuff like: x ^ 1/2 / x ^ 3/4 * x - 4 = x ^ 1/2 / x - (- 1 ) = x ^ 1/2 - ( - 1) = 2 / 2 = 1
then most of us will give up !!
Denis,
Thank you for your help and much needed advice.
I will try to do much better applying the questions in the right proceedure,
and will keep your response and go by that for the next questions.
I apologize for my confusing everyone. (you,, Liz and everyone else who tried to help me).
Thanks again, Helen
Thank you for your help and much needed advice.
I will try to do much better applying the questions in the right proceedure,
and will keep your response and go by that for the next questions.
I apologize for my confusing everyone. (you,
, Liz and everyone else who tried to help me).Thanks again, Helen