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Some Maths Problems

  • Thread starter Bin Qasim
  • Start date
  • #1
Bin Qasim
Hello

I need some help with some maths problems. Actually I am sorta of confused.

My Questions:
1. Rewrite without rational exponents, and simplify, if possible.
a. y^(1/7)...... I am so confused what it is about. I mean I can't do anything with the expressions, can I?
b. (x^2.y^2)^(1/3)...... kinda of similar to above one but wat am I supposed to do with expression....

2. Rewrite with positive exponents.
a. y^(-1/7)........ Am I supposed to just change the sign of power to +ve??

3. Use rational exponents to simplify. Write the answer in radical notation if appropriate.
a. (a^2)^(6)......... :confused:

4. Find domain of f(x)= (x+5)^(1/2). (x+7)^(-1/2) hints badly needed...

I have test tomm so pls reply ASAP. Don't solve it right away but provide me with hints pls.

:smile:
 

Answers and Replies

  • #2
vsage
Sounds like you have difficulty with the language used which will keep you from even starting the problems really.

1a. y^(1/7) is the same as writing the 7th root of y so.. write the 7th root of y a different way.
b. same thing basically as a
2. y^-x = 1/(y^x) right?
3. more power rules. How would you combine the 2 and the 6?
4. Not sure what the "." stands for, but assuming it's multiplication.. simply find the values of x that will result in an imaginary answer. for example, f(-6) is not in the domain.
 
  • #3
870
0
Well, Im not too sure exactly what the question is asking, and what context it is in but I would say that for the first question it is asking you to write it in surd form.

So for 1) a)

[tex]y^{\frac{1}{7}} = \sqrt[7]{y}[/tex]

1)b)

[tex](x^2y^2)^{\frac{1}{3}} = x^{\frac{2}{3}}y^{\frac{2}{3}} = (xy)^{\frac{2}{3}} = \sqrt[3]{xy}^2[/tex]

2)

[tex]y^{-1/7} = \frac{1}{y^{1/7}}[/tex]

3) I dont exactly get the question? There is not much else you can do with this type of question.

[tex](a^2)^6 = a^{12}[/tex]

4) We have

[tex]f(x) = \frac{\sqrt{x+5}}{\sqrt{x+7}}[/tex]

As x approaches -7 from [itex]\infty^-[/itex] it hits an asymptote. So [itex]x\neq -7[/itex] otherwise the denominator is zero. Also, [itex]x \notin (-7,-5)[/itex]. x cannot exist in this interval because

[tex]f(-7) = \frac{\sqrt{-2}}{\sqrt{0}}[/tex] does not exist

[tex]f(-6) = \frac{\sqrt{-1}}{\sqrt{1}} = i \notin \mathbb{R}[/tex]

[tex]f(-5.5) = \frac{\sqrt{-1/2}}{\sqrt{1.5}} = \frac{i}{3}\sqrt{3} \notin \mathbb{R}[/tex]

[tex]f(-5) = 0[/tex]

Hence the domain of [itex]f(x)[/itex] is

[tex] f(x) = (-\infty, -7) \cup (-5,\infty) [/tex]

or

[tex] f(x) \backslash (-7,-5)[/tex]
 
Last edited:
  • #4
Bin Qasim
Thanx guys

my bad vsage..... misunderstood it I guess....

thanx for help guys, vsage and oxymoron..... :smile:

and yeah the third Q is:

Use rational exponents to simplify. Write the answer in radical notation if appropriate.

a^(2/6)

so pls tell me what I am supposed to do ........ thanx :smile:
 
Last edited by a moderator:
  • #5
870
0
Well, to write in radical notation is to write in surd form (with square roots).

So

[tex] a^{\frac{2}{6}} = \sqrt[6]{a}^2[/tex]

By the way, Im using the surd formula:

[tex]x^{\frac{y}{z}} = \sqrt[z]{x}^y[/tex]
 
  • #6
870
0
Also note that with your 4th question, when regarding domains of functions, you must note what field you are working in.

When I did the question I assumed you were working with real numbers.

But if you were working with complex numbers, obviously the function is defined for all [itex]z\in\mathbb{C}\backslash\{-7\}[/tex].
 
  • #7
Bin Qasim
Thanx oxy and vsage that helped me.....

I did my test hope to get an A :smile:

Just one more Q is:

log(x-1)/log(3)= [log(x-2)/log3]-[log(x)/log(3)]

evaluate or simplify the above expression.....

thanx very much appreciated......
 
  • #8
870
0
in your last post Bin Qasim, you say evaluate or simplify. Well, you cant evalute what you have written but you can simplify. This is just a matter of remembering your logarithm laws...

[tex]\frac{\log (x-1)}{\log 3} = \frac{\log (x-2)}{\log 3} - \frac{\log x}{\log 3}[/tex]

[tex]\frac{\log (x-1)}{\log 3} = \frac{\log (x-2) - \log x}{\log 3}[/tex]

However, this should not be an equality since the LHS never equals the RHS. Are you sure there isnt more to this question?
 
  • #9
Bin Qasim
U right oxy, it was to solve for x i forgot that

and one more thing, I go an A in the taht test, I owe u guys for helping me :smile:
 

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