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If 3^(k+1) = (3^9)^3^9 , then k=???

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by DanaJ » Wed Feb 11, 2009 5:58 am
The problem is to untangle (3^9)^3^9 = 3^[9*(3^9)] = 3^[(3^2)*(3^9)] = 3^[3^(2 + 9)]= 3^(3^11). This thing is equal to 3^(k + 1), so k = 3^11 - 1.
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by marcusking » Wed Feb 11, 2009 6:08 am
Did the question provide a list of possible values?

If you are just looking for the answer here's how I solved it (not the most timely method but the best I knew how to do)

we need to find the last 3^9 value. So I started doing it by hand.
3^2=9
3^3=27
3^4=81
3^5=243
3^6=729
3^7=2187
3^8=6561
3^9=19683

so the equation is now 3^(k+1) = (3^9)^19683

since we have (3^9)^19683 given that when you are placing an exponent on an exponent, you can simply multiply the two exponents together 9*19683 = 59049

so now the equation is 3^(k+1) = 3^59049
we can easily solve for k now by subtracting 1 from 59049

to get k = 59048

Once again, this isn't a strong point of mine, but this is my attempt at it. If there is a better solution I would love to know so I too can learn.
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by willbeatthegmat » Wed Feb 11, 2009 6:29 am
ans is k = 3^11 - 1.....marcusking ur ans is in exact numb...i think i shud hve mentioned ans to evade big calculations...thanks so much..danaj knows it all
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by hardik.jadeja » Sat Feb 14, 2009 11:38 am
As we know (a^x)^y = a^(xy).

But is ((a^x)^y)^z equal to a^(xyz)?
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by cramya » Sat Feb 14, 2009 8:10 pm
Good solution DanaJ!
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by cramya » Sat Feb 14, 2009 8:18 pm
As we know (a^x)^y = a^(xy).

But is ((a^x)^y)^z equal to a^(xyz)?
((a^x)^y)^z = (a^x) ^ (yz)
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by hardik.jadeja » Sun Feb 15, 2009 5:25 am
@cramya:

Is it ((a^x)^y)^z = (a^x) ^ (yz) or ((a^x)^y)^z = a^(x*(y^z))???
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by cramya » Sun Feb 15, 2009 6:48 am
Hi Hardik,

I would say ((a^x)^y)^z = (a^x) ^ (yz)

Regards,
CR
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by hardik.jadeja » Sun Feb 15, 2009 7:02 am
@cramya: If you see, DanaJ used ((a^x)^y)^z = a^(x*(y^z)) and i think she is right.
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by cramya » Sun Feb 15, 2009 7:08 am
Try this:

((2^1)^3)^4 = (2 ^1) ^ (3*4) = (2^1) ^12 = 2 ^(1*12) = 2^12 = 8^4 and not 2 ^ (1*3^4)

Hope this helps!
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