Confused about Maths Problems? Get Help and Hints Here!

[Bin Qasim]

Hello

I need some help with some maths problems. Actually I am sort of 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.

 

[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.

 

[Oxymoron]

Well, I am 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 don't 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

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

 

[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 ... thanks

 

[Oxymoron]

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, I am using the surd formula:

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

 

[Oxymoron]

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].[/itex]

 

[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...

 

[Oxymoron]

in your last post Bin Qasim, you say evaluate or simplify. Well, you can't 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 isn't 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