Simplifying Exponent Expression

  • Thread starter Thread starter stonecoldgen
  • Start date Start date
  • Tags Tags
    Simplifying
AI Thread Summary
To simplify the expression (x^(a-1))/x^((3a+4)), apply the property of exponents that states x^m/x^n = x^(m-n). This leads to the simplification: x^((a-1) - (3a+4)). The resulting expression can be further simplified to x^(-2a - 5). It's important to note that this is an expression, not an equation, and the a's cannot cancel out. Understanding these exponent rules is crucial for accurate simplification.
stonecoldgen
Messages
108
Reaction score
0

Homework Statement


simplify:

(x^(a-1))/x^((3a+4))



Homework Equations



i was really dumb to think laws of logarithms would help me, but obviously not...

The Attempt at a Solution



the only thing i know is that the a's can't cancel! i don't know what to do !
 
Physics news on Phys.org
Think about the following:

x^a/x^b=(x^a)(x^-b)

Then,

(x^a)(x^-b)=x^[(a)(-b)]

In the problem you presented think FOIL.
 
stonecoldgen said:

Homework Statement


simplify:

(x^(a-1))/x^((3a+4))
Do you mean this?
\frac{x^{a-1}}{x^{3a+4}}
BTW, I don't mean to nitpick, but this is not an equation. It is an expression. Please don't confuse the two.

czelaya said:
Think about the following:

x^a/x^b=(x^a)(x^-b)

Then,

(x^a)(x^-b)=x^[(a)(-b)]

In the problem you presented think FOIL.
If that's an a times negative b at the end, then that is not correct. We need to use one of the properties of exponents, mainly this one:
\frac{x^m}{x^n}=x^{m-n}
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top