My Math Forum Exponent help

 Algebra Pre-Algebra and Basic Algebra Math Forum

 January 20th, 2016, 06:07 PM #1 Member   Joined: Sep 2015 From: Philadelphia Posts: 36 Thanks: 3 Exponent help Hi I have this one math problem and it's really hard. I got an answer but it's probably wrong. Can someone please solve it and tell me what you get. I want to compare it to my answer Thanks. Here it is: (2(squared) a(to the -1 power) b(to the -3 power) c) the -2 power divided by 8( to the -1 power) ab( to the -5 power) c(to the 0 power) *In the first line, all the terms from 2 to c are in parentheses and the -2 exponent is on the outside of the parentheses
 January 20th, 2016, 06:34 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond Do you mean $\displaystyle \dfrac{\left(2^2\cdot a^{-1}\cdot b^{-3}\cdot c\right)^{-2}}{8^{-1}\cdot(a\cdot b)^{-5} \cdot c^0}$ ? Last edited by skipjack; January 20th, 2016 at 06:49 PM.
 January 20th, 2016, 06:38 PM #3 Member   Joined: Sep 2015 From: Philadelphia Posts: 36 Thanks: 3 No, sorry, it's a little different; how did you type the exponents? The exponent the a and b below the fraction bar are not in the same parentheses but other than that, yes that is what I meant. Please solve if you get a chance. I really appreciate your help. Thanks. Last edited by skipjack; January 20th, 2016 at 06:51 PM.
 January 20th, 2016, 06:44 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond I used LaTeX. Typing $2^{2}$ gives $\displaystyle 2^2$. Typing $\dfrac{4}{3}$ gives $\displaystyle \dfrac{4}{3}$. Typing $\dfrac{4}{3}+2^{2}$ gives $\displaystyle \dfrac{4}{3}+2^{2}$.
 January 20th, 2016, 06:51 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond $\displaystyle \dfrac{\left(2^2\cdot a^{-1}\cdot b^{-3}\right)^{-2}}{8^{-1}\cdot a\cdot b^{-5}\cdot c^0}$ Like that? I get $\displaystyle \dfrac12ab^{11}$.
 January 20th, 2016, 06:56 PM #6 Member   Joined: Sep 2015 From: Philadelphia Posts: 36 Thanks: 3 Thank you so much. The only thing was you forgot the c in the parentheses above the fraction bar but that's okay. Thank you so much. I really appreciate it Thanks from greg1313
January 20th, 2016, 06:57 PM   #7
Global Moderator

Joined: Dec 2006

Posts: 20,281
Thanks: 1965

Quote:
 Originally Posted by hhmathgeek I got an answer . . .

January 20th, 2016, 07:00 PM   #8
Global Moderator

Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,912
Thanks: 1110

Math Focus: Elementary mathematics and beyond
Quote:
 Originally Posted by hhmathgeek The only thing was you forgot the c in the parentheses above the fraction bar...
Yep.

$\displaystyle \dfrac{\left(2^2\cdot a^{-1}\cdot b^{-3}\cdot c\right)^{-2}}{8^{-1}\cdot a\cdot b^{-5}\cdot c^0}=\dfrac12ab^{11}c^{-2}=\dfrac{ab^{11}}{2c^2}$

 January 20th, 2016, 07:03 PM #9 Member   Joined: Sep 2015 From: Philadelphia Posts: 36 Thanks: 3 Thank you so much. My answer was sort of similar but not quite
 January 20th, 2016, 07:05 PM #10 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond Why don't you post your work? We can help you better if you do that.

 Tags exponent

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post The_Ys_Guy Algebra 3 May 1st, 2015 08:21 PM bml1105 Algebra 12 July 27th, 2014 12:40 PM Thinkhigh Calculus 3 March 2nd, 2012 07:42 AM dthomas86 Algebra 7 January 2nd, 2012 01:24 AM alfonso1 Number Theory 17 August 7th, 2007 10:03 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top