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# PS-Exponents

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Senior Manager
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20 Oct 2005, 01:55
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My friend gave me to solve. Don't know the answer choices.

3^6x=8100

(3^(x-1))^3=?

[edited by Hong to correct the question stem as noted below.]
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20 Oct 2005, 03:17
i dont see the trick i can't find the path
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20 Oct 2005, 03:50
I don't know the official answer but I still gave it a try. Just don't know how far it is correct.

Since 3^6x=8100

3^6 * 3^x=8100

3^x=8100/3^6
=8100/243
=100/3

Substituting value of 3^x in the equation (3^x-1)^3

((100/3)/(1/3))^3

(100/9)^3.
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20 Oct 2005, 03:55
getzgetzu wrote:
My friend gave me to solve. Don't know the answer choices.

3^6x=8100

(3^x-1)^3=?

use the formula (a-b)^3= a^3 - 3a^2*b + 3 a*b^2 - b^3

(3^x-1)^3= 3^(3x) - 3*3^(2x) + 3*3^x -1
=sqrt(3^(6x))- 3* cubic root of 3^(6x) + 3* 6th root of 3^(6x)-1
= 90+ 60.25 ( rounded) + 13.45 (rounded)-1
= 162.7
Senior Manager
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20 Oct 2005, 03:57
Can somebody help me where did I wrong? Please.
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20 Oct 2005, 03:58
see my explanation above ...hope this helps
Senior Manager
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20 Oct 2005, 04:09
I am extremely sorry for the typo

3^6x=8100

(3^(x-1))^3=?

So we cannot use (a-b)^3 formula here. I wish we had MathTypeâ„¢, we would have avoided all such errors.
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20 Oct 2005, 04:16
getzgetzu wrote:
I am extremely sorry for the typo

3^6x=8100

(3^(x-1))^3=?

So we cannot use (a-b)^3 formula here. I wish we had MathTypeâ„¢, we would have avoided all such errors.

It's even easier!

(3^(x-1))^3= 3^(3*(x-1)) = 3^(3x-3)= (3^3x)/ 3^3= sqrt(3^(6x))/27=90/27= 10/3
20 Oct 2005, 04:16
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