My Math Forum Can't figure out the k exponent variable?

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September 8th, 2014, 12:40 PM   #11
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Quote:
 Originally Posted by DecoratorFawn82 5^11 * (1/5^k) = 5 5^11 / (1 * 5^k) = 5 5^11 / 5^k = 5 5^11-k = 5^1 k = 10
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
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Quote:
 Originally Posted by raju 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
No, raju, almost all of your post is wrong. I have already mentioned to you
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^(n-m) 3)x^-n = 1 / x^n 4)x^m = x^n i.e., m=n
September 10th, 2014, 08:40 AM   #16
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Quote:
 Originally Posted by aurel5 I still try that: $\displaystyle \dfrac{5^{11}}{5^k}=5 \\\;\\ \dfrac{5\cdot5^{10}}{5^k}=5|_{:5} \\\;\\ \dfrac{5^{10}}{5^k}=1 \\\;\\ 5^k = 5^{10} \\\;\\ k = 10 .$
shorter way :
$\displaystyle \frac{5^{11}}{5^{k}} = 5$

using $\displaystyle \frac{a^{m}}{a^{n}} = a^{m-n}$

$\displaystyle 5^{11-k} = 5^{1}$

$\displaystyle \text{solving for powers , 11 - k = 1}$

$\displaystyle \text{-k = -10 or k = 10}$

 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: 21,029 Thanks: 2259 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, 5^11-k should be 5^(11-k) or $\displaystyle 5^{11-k}$ 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^(n-n) = (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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