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[SOLVED] Help with dividing an exponent

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
MDS1005's question from Math Help Forum,


I'll keep this short. My understanding is dividing an exponent works as follows:

x^2/x^-5=x^7

The exponent rules state that you subtract the exponent in the denominator from the exponent in the numerator if they have the same base. 2 - (-5) = 7

Now, my question. Maybe I'm wrong, or maybe the answer I am looking at is wrong. I need to know which.

5x^2/10x^-5

My answer: 1/2x^7
Answer I was given: .5/x^5

The same source which provided this answer confirms my first idea of x^2/x^-5 = x^7. What changes when 5/10 is added onto the whole deal?
Hi MDS1005,

I hope you meant, that your answer is, \(\dfrac{1}{2}x^7\). If that is the case, your answer is correct and the given answer is wrong.


\[\frac{5x^{2}}{10x^{-5}}=\frac{1}{2}x^{2-(-5)}=\frac{1}{2}x^7\]
 
Last edited:

MDS1005

New member
May 27, 2012
1
Thanks. I knew that I was correct. I actually lost a half hour of sleep last night trying to figure out how the answer I was given could possibly be correct (Wondering). And now, thanks to your recommendation I now have a new forum to ask me questions in. Thanks for that Sudharaka. I'm sure I'll have many many more questions to come, seeing as I just registered for school as a physics major and still have questions like this. Physics is going to be quite an undertaking after 10 years of not doing any math in a formal setting.
 

checkittwice

Member
Apr 3, 2012
37
MDS1005,

you must use grouping symbols because of the Order of Operations.


So, then you would have:

5x^2/(10x^-5)


Better:

(5x^2)/(10x^-5)



Or, as:

[tex]5x^2/(10x^{-2})[/tex]



Or, better:

[tex](5x^2)/(10x^{-2})[/tex]