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exponents.....

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exponents.....

by vaswani.sharan » Sun Apr 15, 2012 10:53 am
Hey, just getting my math feet back under me, I am not so bad at exponents, but this question is bugging me. Not sure how I could quickly solve it.
can some one show me a fast way?

(5^21)(4^11)=2x10^n
n=21
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by shubham_k » Sun Apr 15, 2012 11:00 am
We see "5" and "4" on the left side and "10" on the right side.. somehow we want to make the bases the same.

So, let's start by turning the "4" into "2^2":

(5^21)((2^2)^11) = 2*10^n

(5^21)(2^22) = 2*10^n

Now, we want to be able to simplify the left side, but we can only do so if the exponents are equal, so:

(5^21)(2^21)(2^1) = 2*10^n

if we divide both sides by 2, we get:

(5^21)(2^21) = 10^n

and since the exponents are equal on the left side, we can say:

(5*2)^21 = 10^21

and finally

10^21 = 10^n

so 21 = n.

Not a difficult one i guess.....
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by Shalabh's Quants » Sun Apr 15, 2012 12:10 pm
vaswani.sharan wrote:Hey, just getting my math feet back under me, I am not so bad at exponents, but this question is bugging me. Not sure how I could quickly solve it.
can some one show me a fast way?

(5^21)(4^11)=2x10^n
n=21
Lets break it into prime factors...

(5^21)(4^11)=2x10^n

5^21*(2^2)^11=2*(2.5)^n

5^21*2^22 = 2*2^n.5^n

5^21*2^22 = 2^(n+1).5^n

Either equate exponents of 2 or 5. Both gives n = 21.
Shalabh Jain,
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by Anurag@Gurome » Sun Apr 15, 2012 5:21 pm
vaswani.sharan wrote:Hey, just getting my math feet back under me, I am not so bad at exponents, but this question is bugging me. Not sure how I could quickly solve it.
can some one show me a fast way?

(5^21)(4^11)=2x10^n
n=21
(5^21)(4^11) = 2 x 10^n
(5^21)(2^2)^11 = 2 x (2 * 5)^n
(5^21)(2^22) = 2 x 2^n * 5^n
(5^21)(2^22) = 2^(n + 1) * 5^n
Now since the bases are the same, so exponents will also be the same on both sides.
So, n = 21
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