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Source: — Problem Solving |
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by mike22629 » Thu Apr 02, 2009 12:04 pm
First find x:
3^(6x) = 8100
Factor 8100 = 3*3*3*3*5*5*2*2 = 3^4*5^2*2^2

So simplify 3^(6x) = 3^4*5^2*2^2
3^x = 10

With next problem use exponent of exponent law
(3^(x-1)^3 = 3^(3x-3) = (3^(3x))/ (3^3) = (3^x)/3

Since 3^x = 10

Answer 10/3
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by moutar » Thu Apr 02, 2009 1:14 pm
(3^(x-1))^3

= 3^(3x-3)

= 3^3x /3^3

= ((3^6x)^1/2)/27

= (8100^1/2)/27

= 90/27

= 10/3 = D

Much nicer.
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by vittalgmat » Thu Apr 02, 2009 1:23 pm
moutar wrote:(3^(x-1))^3

= 3^(3x-3)

= 3^3x /3^3

= ((3^6x)^1/2)/27

= (8100^1/2)/27

= 90/27

= 10/3 = D

Much nicer.
neat!!
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by gmat740 » Thu Apr 02, 2009 5:07 pm
I didn't got what the problem is saying:

when we say 2^3
we mean 8
and neither 6, nor 9 (3^2) correct??


So,
(3^(x-1))^3

= 3^(3x-3)

So why 3 in the right hand exponent side is being multiplied instead of putting it into power??
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by mike22629 » Thu Apr 02, 2009 6:00 pm
Nice Moutar! That is both quicker and simpler. Thanks
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by Vemuri » Tue Apr 07, 2009 8:35 pm
moutar wrote:(3^(x-1))^3

= 3^(3x-3)

= 3^3x /3^3

= ((3^6x)^1/2)/27

= (8100^1/2)/27

= 90/27

= 10/3 = D

Much nicer.
Really cool !!! Just shows that we should be open for different perspectives while solving the question. I just hope I can think like this on the D day :-)
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by kaf » Wed Apr 08, 2009 2:29 am
moutar wrote:(3^(x-1))^3

= 3^(3x-3)

= 3^3x /3^3

= ((3^6x)^1/2)/27

= (8100^1/2)/27

= 90/27

= 10/3 = D

Much nicer.
I have serious problems with exponent questions.

Moutar can you please explain how

= 3^3x /3^3 transformed to become

= ((3^6x)^1/2)/27

thanks in advance
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by moutar » Wed Apr 08, 2009 2:41 am
(a^x)^y = a^(x*y)

Therefore (3^6x)^1/2 = 3^3x

and 3^3 = 27.

That OK?
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by kaf » Wed Apr 08, 2009 3:47 am
Thanks a lot Moutar

its clear now
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by gmat740 » Wed Apr 08, 2009 6:04 am
Oh!!
I got it

Thanks a Lot Moutar
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