MHB Help with dividing an exponent

  • Thread starter Thread starter Sudharaka
  • Start date Start date
  • Tags Tags
    Exponent
Click For Summary
Dividing exponents involves subtracting the exponent in the denominator from the exponent in the numerator when the bases are the same. In the case of 5x^2/10x^-5, the correct simplification leads to 1/2x^7, confirming MDS1005's initial understanding. The confusion arose from the addition of the coefficients 5 and 10, which must be simplified separately. Proper use of grouping symbols is emphasized to clarify the order of operations. Overall, MDS1005 is correct in their calculations and has found a supportive forum for future questions.
Sudharaka
Gold Member
MHB
Messages
1,558
Reaction score
1
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:
Mathematics news on Phys.org
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.
 
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:

5x^2/(10x^{-2})
Or, better:

(5x^2)/(10x^{-2})
 
Thread 'Erroneously  finding discrepancy in transpose rule'
Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K