Wikipedia:Reference desk/Archives/Mathematics/2010 September 13

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September 13

exponents

Hi, I am supposed to solve this and show it in exponential form

(3x)^2 (9x^-1)

So, so far I have done is:

9x^2 (9x^-1)

9x^2 (1/9x)

x (1/1)

x

But I have a feeling I didn't do this right, since I'm supposed to answer it in an "exponential" form. Can anyone help please? --174.5.131.59 (talk) 04:12, 13 September 2010 (UTC)

Is the expression (3x)²(9x)^-1 or (3x)²(9x^-1)? If it's the second one, then you should check carefully what you did. Rckrone (talk) 04:40, 13 September 2010 (UTC)

Thanks for showing me the superscripts! It is the second one, (3x)²(9x^-1) I did check carefully over and over but the solution seems to be escaping me. Can you give me some tips please? --174.5.131.59 (talk) 05:18, 14 September 2010 (UTC)

In 9x^-1, the exponent is only applying to the x, not to the 9. In your second line, you applied it to the 9.--203.97.79.114 (talk) 05:22, 14 September 2010 (UTC)

Yes ab^c means a(b^c), not (ab)^c. E=mc² means E=m(c²)=c²m. It is common practice to put the constant, which here is c², before the variable, which here is m, but Einstein did not do that in this case, and everybody follow Einstein and write E=mc². Note the silly exception: cm² means square centimeter, not centi squaremeter, so cm² = (cm)². Bo Jacoby (talk) 06:03, 14 September 2010 (UTC).

Inverse function

How do I invert $y={\frac {1-{\sqrt {x}}}{1+{\sqrt {x}}}}$

to get $y^{-1}=({\frac {1-x}{1+x}})^{2}$ —Preceding unsigned comment added by 118.208.129.11 (talk) 08:58, 13 September 2010 (UTC)

First step is to swap the x's and y's. Then make y the subject of the formula by multiplying out by the denominator and grouping like terms. Can't give any more hints than that without actually doing it for you. Ask if you're still stuck. Zunaid 10:21, 13 September 2010 (UTC)

Be careful with notation. y⁻¹ usually means the reciprocal of y. But apparently that's not what you mean here. Michael Hardy (talk) 03:53, 14 September 2010 (UTC)

Math

I'm not so good at maths. It math something you learn, or something you're always good at? Mediaiiwk (talk) 16:09, 13 September 2010 (UTC)

It is something you learn, and then you enjoy it, and so you learn even more, and so on. Be patient and have fun doing math, then you too will be good at it. Bo Jacoby (talk) 16:31, 13 September 2010 (UTC).

Agree with Bo: the biggest obstacle to learning math, for most people, is that they are afraid they won't be good at it. You just have to remember that no one is good at it to start, but if you keep working at it you will get it, and the moment you get it is a really satisfying feeling. --Ludwigs2 19:20, 13 September 2010 (UTC)

It is similar to the situation with English literature. There are many writers who try very hard and go to all sorts of courses about good writing but only a small percentage write anything you'd want to read. And even if you're content to read novels rather than try and write them there are some that can be quite heavy going and yet are acknowledged as great literature. And then there's people who don't read. And then there's people who can't read and even with good help have trouble getting what they want in a supermarket. There's quite a few good books in popular maths that are accessible to most people with some interest like novels are to those who like reading. And like writing reasonable English so you can say what you think is helped by a bit of training so most people could do better with their maths and get their way past interest rates, builders estimates,, sizing a room for painting and suchlike tasks with reasonable ease. Dmcq (talk) 10:11, 14 September 2010 (UTC)