September 8th, 2014, 12:40 PM  #11 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,943 Thanks: 1134 Math Focus: Elementary mathematics and beyond  That's it. You can skip the second step though.
Last edited by greg1313; September 8th, 2014 at 12:51 PM. 
September 8th, 2014, 01:13 PM  #12 
Senior Member Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177  I still try that: $\displaystyle \dfrac{5^{11}}{5^k}=5 \\\;\\ \dfrac{5\cdot5^{10}}{5^k}=5_{:5} \\\;\\ \dfrac{5^{10}}{5^k}=1 \\\;\\ 5^k = 5^{10} \\\;\\ k = 10 .$ 
September 9th, 2014, 09:54 AM  #13 
Newbie Joined: Sep 2014 From: hydedrabad Posts: 22 Thanks: 4  algebra problem
Remember this formula: (a)^m (a)^n=(a)^m+n Explanation: when the bases in a exponent are equal and are in a multiplying operation, you can do adding of exponents as shown in the above formula. According to this formula, the above question changes to 3^k+4 =3^9 As the bases are equal, powers are equal K+4=9 K=5 
September 9th, 2014, 03:20 PM  #14  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
about the necessity of grouping symbols in at least one other thread because of the Order of Operations. 1) Parentheses are unneeded around the bases at the top of the page. 2) Grouping symbols are needed around the exponent that is a binomial wherever that happens. 3) Once you're in lower case for that letter, you stay that way. A "K" does not represent the same variable as a "k" does. Here is a possible amendment: (a^m)*(a^n) = a^(m + n) 3^(k + 4) = 3^9 k + 4 = 9 k = 5  
September 10th, 2014, 07:33 AM  #15 
Newbie Joined: Sep 2014 From: us Posts: 10 Thanks: 1 
The following exponent laws are useful to understand 1)x^n * x^m = x^(n+m) 2)x^n / x^m = x^(nm) 3)x^n = 1 / x^n 4)x^m = x^n i.e., m=n 
September 10th, 2014, 08:40 AM  #16  
Senior Member Joined: Jul 2014 From: भारत Posts: 1,178 Thanks: 230  Quote:
$\displaystyle \frac{5^{11}}{5^{k}} = 5$ using $\displaystyle \frac{a^{m}}{a^{n}} = a^{mn}$ $\displaystyle 5^{11k} = 5^{1}$ $\displaystyle \text{solving for powers , 11  k = 1}$ $\displaystyle \text{k = 10 or k = 10}$  
September 10th, 2014, 10:16 AM  #17 
Newbie Joined: Sep 2014 From: hydedrabad Posts: 22 Thanks: 4  exponent k value
For doing this problem, you must know these 2 formulas. Formula 1: (a)^m*(a)^n= (a)^ m+n Formula2: if two exponents are in the form a^x=a^y Then x=y 
September 10th, 2014, 03:59 PM  #18 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133 
In your answers that contain a "5/5 = 1" step, that step doesn't really make sense, and hence doesn't explain why your next line is correct. In your penultimate line, 5^11k should be 5^(11k) or $\displaystyle 5^{11k}$ so that k is part of the exponent. Note that p*(1/q) = (p*1)/q = p/q regardless of what p and q are. Alternatively, as 5^0 = 5^(nn) = (5^n)/(5^n) = 1 and 5^1 = 5, 5^11 * (1/5^k) = 5 is equivalent to 5^11 * 5^0 / 5^k = 5^1, which implies 11 + 0  k = 1, and so k = 10. 

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