I'm up to the exponents chapter of the book and I'm basically done with it, but I have concerns about 6 things in the chapter. I hope that I'm allowed to post that many on here without having to start a new thread. Here goes…
1) Are these expressions equivalent? The book seems to indicate that they are when it does the quick law of exponents section.
\(\displaystyle \frac{x^a}{x^b}=x^{a-b}=\frac{1}{x^{b-a}}\)
I'm assuming that they are all equivalent. The equivalency holds for at least a few of the values I've plugged in.
2) When doing a problem like this:
\(\displaystyle (x+y)^0\)
The answer will come to 1. That's fine, and perfectly understandable to me, but at the same time I keep getting the temptation to make the answer 2. The reasoning goes like this: \(\displaystyle x^0=1\) and \(\displaystyle y^0=1\), and 1+1=2. I keep thinking this because earlier in the book there's a pattern given for the law of exponents that goes like this \(\displaystyle (xy)^a=x^ay^a\). Which makes it seem as if the law of exponents is to distribute the exponent to each item inside the parentheses. Is that a reasonable interpretation? Should I just assume that things are different for addition/subtraction in parentheses than it is for multiplication/division, and be done with this weird temptation?
3) The book is asking me to simplify this problem:
\(\displaystyle \frac{px^{p-1}}{q(x^{p/q})^{q-1}}\)
I didn't know what to do with this one. The answer given is, \(\displaystyle \frac{px^{p/q-1}}{q}\). Does the \(\displaystyle ^{p/q \times q-1}\) part turn into \(\displaystyle ^{p-p/q}\)? If so, I don't really know how to divide that into \(\displaystyle p-1\). It seems that the \(\displaystyle px\) part divided by what will turn into \(\displaystyle qx\) will just give me \(\displaystyle p/q\). But the final answer has \(\displaystyle px/q\). Why does the numerator keep its \(\displaystyle x\) while the denominator loses everything? Moving on…
4) This part wanted me to remove common factors. Here's the example problem:
\(\displaystyle 3(x+3)^2(3x-1)^{1/2}+(x+3)^3(\frac{1}{2})(3x-1)^{-1/2}(3)\)
I turned this into
\(\displaystyle \frac{6}{2}(x+3)^2(3x-1)^{1/2}+\frac{3}{2}(x+3)^3(3x-1)^{-1/2}\)
and from this I concluded
\(\displaystyle \frac{3}{2}(x+3)^2(3x-1)^{1/2}\times2+(x+3)(3x-1)^{-1}\)
which is
\(\displaystyle \frac{3}{2}(x+3)^2(3x-1)^{1/2}(\frac{5+x}{3x-1})\)
That's gone off in the wrong direction I can see by looking at it, and by comparing it to the answer that the book gave. The book even gave a different common factor
\(\displaystyle \frac{3}{2}(x+3)^2(3x-1)^{-1/2}\)
Which I found strange as I figured the object was to get the greatest common factor, and the positive exponent is greater than the negative one. Help?
5) From that same "remove common factors" problem set:
\(\displaystyle -\frac{3}{2}(4x^2-1)^{-5/2}(8x)(1+x^2)^{2/3}+(4x^2-1)^{-3/2}(\frac{2}{3})(1+x^2)^{-1/3}(2x)\)
I started by first turning this into
\(\displaystyle -\frac{24x}{2}(4x^2-1)^{-5/2}(1+x^2)^{2/3}+\frac{4x}{3}(4x^2-1)^{-3/2}(1+x^2)^{-1/3}\)
then I made common denominators with the lowest common multiple of 2 and 3 and turned the sum into
\(\displaystyle -\frac{72x}{6}(4x^2-1)^{-5/2}(1+x^2)^{2/3}+\frac{8x}{6}(4x^2-1)^{-3/2}(1+x^2)^{-1/3}\)
Finding the greatest common factor I converted the sum into
\(\displaystyle \frac{8x}{6}(4x^2-1)^{-5/2}(1+x^2)^{2/3} \times -\frac{9}{6}(4x^2-1)^{2/2}(1+x^2)^{-3/3}\)
which simplifies to
\(\displaystyle \frac{8x}{6}(4x^2-1)^{-5/2}(1+x^2)^{2/3} \times -\frac{9}{6}(4x^2-1)(1+x^2)^{-1}\)
and then
\(\displaystyle \frac{8x}{6}(4x^2-1)^{-5/2}(1+x^2)^{2/3} \times (-\frac{36x^2}{6}+\frac{9}{6})\times \frac {1}{1+x^2}\)
and then
\(\displaystyle \frac{8x}{6}(4x^2-1)^{-5/2}(1+x^2)^{2/3} \times (-6x^2+\frac{9}{6})\times \frac {1}{1+x^2}\).
At that point I stopped going as that looks completely wrong. In writing all this out to explain to others my confusion, I think I found my way out of this. What needs to be done is to either 1. take an expression from both sides of the + sign in order to make sure that I end up with something that cancels out. Or that I 2. take the negative exponent when I have the option, so that I end up with a positive one later. Had I chosen either option I would've ended up with the answer that the book gave (I think). Is that the right conclusion to draw?
6) This one was a "simplify and write in scientific notation" problem:
\(\displaystyle \frac {(3 \times 10^{-5})(6 \times 10^{-3})^3}{(9 \times 10^{-12})^2}\)
I thought it would be straightforward enough if I just did it like this
\(\displaystyle \frac {(3 \times 10^{-5})(216 \times 10^{-9})}{(81 \times 10^{-24})}\) = \(\displaystyle \frac {642 \times 10^{-14}}{81 \times 10^{-24}}\) = \(\displaystyle \frac {642 \times 10^{24}}{81 \times 10^{14}}\) = \(\displaystyle 7.\overline{925} \times 10^{10}\)
But apparently the answer is actually
\(\displaystyle 8 \times 10^{10}\)
I don't think I've ever gotten an answer so close to the correct one by doing the wrong thing before. What did I do wrong here?
Thanks to anyone for the help they can give, and I hope that's not too many questions in one post.
1) Are these expressions equivalent? The book seems to indicate that they are when it does the quick law of exponents section.
\(\displaystyle \frac{x^a}{x^b}=x^{a-b}=\frac{1}{x^{b-a}}\)
I'm assuming that they are all equivalent. The equivalency holds for at least a few of the values I've plugged in.
2) When doing a problem like this:
\(\displaystyle (x+y)^0\)
The answer will come to 1. That's fine, and perfectly understandable to me, but at the same time I keep getting the temptation to make the answer 2. The reasoning goes like this: \(\displaystyle x^0=1\) and \(\displaystyle y^0=1\), and 1+1=2. I keep thinking this because earlier in the book there's a pattern given for the law of exponents that goes like this \(\displaystyle (xy)^a=x^ay^a\). Which makes it seem as if the law of exponents is to distribute the exponent to each item inside the parentheses. Is that a reasonable interpretation? Should I just assume that things are different for addition/subtraction in parentheses than it is for multiplication/division, and be done with this weird temptation?
3) The book is asking me to simplify this problem:
\(\displaystyle \frac{px^{p-1}}{q(x^{p/q})^{q-1}}\)
I didn't know what to do with this one. The answer given is, \(\displaystyle \frac{px^{p/q-1}}{q}\). Does the \(\displaystyle ^{p/q \times q-1}\) part turn into \(\displaystyle ^{p-p/q}\)? If so, I don't really know how to divide that into \(\displaystyle p-1\). It seems that the \(\displaystyle px\) part divided by what will turn into \(\displaystyle qx\) will just give me \(\displaystyle p/q\). But the final answer has \(\displaystyle px/q\). Why does the numerator keep its \(\displaystyle x\) while the denominator loses everything? Moving on…
4) This part wanted me to remove common factors. Here's the example problem:
\(\displaystyle 3(x+3)^2(3x-1)^{1/2}+(x+3)^3(\frac{1}{2})(3x-1)^{-1/2}(3)\)
I turned this into
\(\displaystyle \frac{6}{2}(x+3)^2(3x-1)^{1/2}+\frac{3}{2}(x+3)^3(3x-1)^{-1/2}\)
and from this I concluded
\(\displaystyle \frac{3}{2}(x+3)^2(3x-1)^{1/2}\times2+(x+3)(3x-1)^{-1}\)
which is
\(\displaystyle \frac{3}{2}(x+3)^2(3x-1)^{1/2}(\frac{5+x}{3x-1})\)
That's gone off in the wrong direction I can see by looking at it, and by comparing it to the answer that the book gave. The book even gave a different common factor
\(\displaystyle \frac{3}{2}(x+3)^2(3x-1)^{-1/2}\)
Which I found strange as I figured the object was to get the greatest common factor, and the positive exponent is greater than the negative one. Help?
5) From that same "remove common factors" problem set:
\(\displaystyle -\frac{3}{2}(4x^2-1)^{-5/2}(8x)(1+x^2)^{2/3}+(4x^2-1)^{-3/2}(\frac{2}{3})(1+x^2)^{-1/3}(2x)\)
I started by first turning this into
\(\displaystyle -\frac{24x}{2}(4x^2-1)^{-5/2}(1+x^2)^{2/3}+\frac{4x}{3}(4x^2-1)^{-3/2}(1+x^2)^{-1/3}\)
then I made common denominators with the lowest common multiple of 2 and 3 and turned the sum into
\(\displaystyle -\frac{72x}{6}(4x^2-1)^{-5/2}(1+x^2)^{2/3}+\frac{8x}{6}(4x^2-1)^{-3/2}(1+x^2)^{-1/3}\)
Finding the greatest common factor I converted the sum into
\(\displaystyle \frac{8x}{6}(4x^2-1)^{-5/2}(1+x^2)^{2/3} \times -\frac{9}{6}(4x^2-1)^{2/2}(1+x^2)^{-3/3}\)
which simplifies to
\(\displaystyle \frac{8x}{6}(4x^2-1)^{-5/2}(1+x^2)^{2/3} \times -\frac{9}{6}(4x^2-1)(1+x^2)^{-1}\)
and then
\(\displaystyle \frac{8x}{6}(4x^2-1)^{-5/2}(1+x^2)^{2/3} \times (-\frac{36x^2}{6}+\frac{9}{6})\times \frac {1}{1+x^2}\)
and then
\(\displaystyle \frac{8x}{6}(4x^2-1)^{-5/2}(1+x^2)^{2/3} \times (-6x^2+\frac{9}{6})\times \frac {1}{1+x^2}\).
At that point I stopped going as that looks completely wrong. In writing all this out to explain to others my confusion, I think I found my way out of this. What needs to be done is to either 1. take an expression from both sides of the + sign in order to make sure that I end up with something that cancels out. Or that I 2. take the negative exponent when I have the option, so that I end up with a positive one later. Had I chosen either option I would've ended up with the answer that the book gave (I think). Is that the right conclusion to draw?
6) This one was a "simplify and write in scientific notation" problem:
\(\displaystyle \frac {(3 \times 10^{-5})(6 \times 10^{-3})^3}{(9 \times 10^{-12})^2}\)
I thought it would be straightforward enough if I just did it like this
\(\displaystyle \frac {(3 \times 10^{-5})(216 \times 10^{-9})}{(81 \times 10^{-24})}\) = \(\displaystyle \frac {642 \times 10^{-14}}{81 \times 10^{-24}}\) = \(\displaystyle \frac {642 \times 10^{24}}{81 \times 10^{14}}\) = \(\displaystyle 7.\overline{925} \times 10^{10}\)
But apparently the answer is actually
\(\displaystyle 8 \times 10^{10}\)
I don't think I've ever gotten an answer so close to the correct one by doing the wrong thing before. What did I do wrong here?
Thanks to anyone for the help they can give, and I hope that's not too many questions in one post.
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