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Expo Problem

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by raju232007 » Fri Oct 03, 2008 12:09 pm
Pls check your question properly...

3^6x=8100
Taking square root on both the sides
3^3x=90...1

We re asked to find (3^(x-1))^2
i.e 3^(3x-3)
3^3x/3^3
From 1 we know that 3^3x=90
therefore 90/27=10/3

Ans is D..10/3

Correct me if i am wrong...
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Re: Expo Problem

by Morgoth » Fri Oct 03, 2008 12:29 pm
smallsorrow wrote:If 3^6x = 8100, what is the value of (3x-1)^3?
A)90
B)30
C)10
D)10/3
E)10/9
Are you sure that this part "(3x-1)^3" is correctly posted

3^(x-1)^3

(x-1)^3 = x^3 - 1^3 - 3x(x-1) = x^3 - 1 -3x^2 + 3x

3^[x^3 - 1 -3x^2 + 3x] = [3^x^3 * 3^3x] / [3^1 *3^3x^2]


= 3^3x * 3^x / 3

3^6x = 8100
(3^3x)^2 = 90^2
3^3x = 90

90*3^x /3 = 30*3^x

if 90 is the answer, x=1, which is not possible. 3^6x = 8100
if 30 is the answer, x=0, which is not possible.
if 10 is the answer, x=-1, which is not possible.
if 10/3 is the answer, x=-2, which is not possible.
if 10/9 is the answer, x=-3, which is not possible.

either the question is wrong or I am missing something.
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by smallsorrow » Fri Oct 03, 2008 1:01 pm
Actually it says ^3 in the end

might be a printing mistake... cause I havnt found a solution either...
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