Some Maths Problems

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).........

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.

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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.

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)

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

1)b)

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

2)

$$y^{-1/7} = \frac{1}{y^{1/7}}$$

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

$$(a^2)^6 = a^{12}$$

4) We have

$$f(x) = \frac{\sqrt{x+5}}{\sqrt{x+7}}$$

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

$$f(-7) = \frac{\sqrt{-2}}{\sqrt{0}}$$ does not exist

$$f(-6) = \frac{\sqrt{-1}}{\sqrt{1}} = i \notin \mathbb{R}$$

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

$$f(-5) = 0$$

Hence the domain of $f(x)$ is

$$f(x) = (-\infty, -7) \cup (-5,\infty)$$

or

$$f(x) \backslash (-7,-5)$$

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Bin Qasim
Thanx guys

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

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

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

Last edited by a moderator:
Well, to write in radical notation is to write in surd form (with square roots).

So

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

By the way, Im using the surd formula:

$$x^{\frac{y}{z}} = \sqrt[z]{x}^y$$

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].

Bin Qasim
Thanx oxy and vsage that helped me.....

I did my test hope to get an A

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......

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...

$$\frac{\log (x-1)}{\log 3} = \frac{\log (x-2)}{\log 3} - \frac{\log x}{\log 3}$$

$$\frac{\log (x-1)}{\log 3} = \frac{\log (x-2) - \log x}{\log 3}$$

However, this should not be an equality since the LHS never equals the RHS. Are you sure there isnt more to this question?

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