# Not looking for answer, just whining. For now.

• Sep 1st 2012, 04:35 PM
kevro2000
Not looking for answer, just whining. For now.
Attachment 24662

I'm supposed 2 simplify this, and according to my Barron's Precalc The EZ way, the answer is B^n.
So far, I've not gotten there, but hopefully I will soon. So far, I've not had issues with this problem set, but for some reason, this one and the next one (same type of problem) are stumping me.

I will continue working with this and hopefully will get it, soon. I suspect that I need to start with the paranthetical cube in the numerator, but I feel like I'm 15 yrs old again (I'm in my 40s ☺) staring at my text book. Good grief, and this was my idea to try this book on for size.

K
• Sep 1st 2012, 05:18 PM
Plato
Re: Not looking for answer, just whining. For now.
Quote:

Originally Posted by kevro2000

$\frac{{{b^{{{\left( {2n + 1} \right)}^3}}}}}{{{b^n} \cdot {b^{4n + 1}}}} = \frac{{{b^{8{n^3} + 12{n^2} + 6n + 1}}}}{{{b^{5n + 1}}}} = {b^{8{n^3} + 12{n^2} + n}}$
• Sep 6th 2012, 06:09 PM
kevro2000
Re: Not looking for answer, just whining. For now.
Plato, thank you: among my multi-solutions that I've come up with, your is the most logical, but not anything like the 2 or 3 that I came up with, and certainly not the "B^n" the Barron's EZ precalc gives as the answer. I'm going to study how to deal with exponents, and then come back to this section of questions in the book.

Thanks, much,
Kevin
• Sep 6th 2012, 06:21 PM
SworD
Re: Not looking for answer, just whining. For now.
Please note: a^(b^c) is NOT the same as (a^b)^c. Exponentiation is NOT associative.

Are you sure that the initial equation isn't supposed to mean $\left (b^{2n+1} \right )^3$ instead? If it does, then the answer would indeed be b^n.

$\frac{\left (b^{2n+1} \right )^3}{b^n \cdot b^{4n+3}}$

Note that (a^b)^c = a^(b*c)

= $\frac{b^{6n+3}}{b^n \cdot b^{4n+3}}$

= $\frac{b^{6n+3}}{b^{5n+3}}$

= $b^{6n + 3 - 5n - 3} = b^n$

The fundamental property of exponents, valid everywhere, is that $x^{a+b} = x^a \cdot x^b$. So when b is negative, this leads to $x^{a-b} = \frac{x^a}{x^b}$. Make use of this property ;)
• Sep 7th 2012, 11:46 AM
kevro2000
Re: Not looking for answer, just whining. For now.
I'll double-check the book tonite, and make sure I've entered and am showing it correctly, as in the opening thread. I'm pretty sure I copied it and got it right, and i used MathOMir to turn it into an image: but just in case, I want to be sure. I understand both examples you two have given, and will look for more samples online to play with. My Barrons book only has two in this format (at least in this section of the book.)
• Sep 8th 2012, 11:23 AM
kevro2000
Re: Not looking for answer, just whining. For now.
SworD, you are correct: I mis-keyed the equation. I see on my paper-work where I was first attempting it, I copied it correctly, but still worked it incorreclty. A few days later, on paper, I had copied it again, thinking I'd start over on it and do better, but I miswrote it down. From that, I used mathomir, and that is why the problem appears as such at the start of the thread.

► So, you are correct, the numerator is (b^2n+1)^3 All in parantheses, cubed. ◄
Armed with the above comments, I will try this problme again, on my own. Thanks for the help, folks.
• Sep 9th 2012, 01:39 PM
kevro2000
Re: Not looking for answer, just whining. For now.
Yay!!! Got it and the next (similar problem) in the book. Moving on to the word problems, now. Onward and forward...........I hope.
• Sep 11th 2012, 05:04 PM
HallsofIvy
Re: Not looking for answer, just whining. For now.
For me it's more often "Onward and aroundward"!