SOLUTION: I can't solve this problem. Please help me. (3^x)(9^x-1) = 27^1-x My solution: 27^2x-1 = 27^1-x I don't know what to do next. Thank You.

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Question 881970: I can't solve this problem. Please help me.
(3^x)(9^x-1) = 27^1-x
My solution:
27^2x-1 = 27^1-x
I don't know what to do next. Thank You.

Found 2 solutions by lwsshak3, MathTherapy:
Answer by lwsshak3(11628)   (Show Source): You can put this solution on YOUR website!
(3^x)(9^x-1) = 27^1-x
(3^x)(3^2(x-1) = 3^3(1-x)
(3^x)(3^(2x-2) = 3^(3-3x)
3^(3x-2) = 3^(3-3x)
3x-2=3-3x
6x=5
x=5/6
Check:
3^(5/6)*9^(-1/6)=27^1/6
(3^(5/6))/9^(1/6)=3^(3/6)
(3^(5/6))/3^(2/6)=3^(3/6)
3^(3/6)=3^(3/6)
check ok
Answer by MathTherapy(10555)   (Show Source): You can put this solution on YOUR website!
I can't solve this problem. Please help me.
(3^x)(9^x-1) = 27^1-x
My solution:
27^2x-1 = 27^1-x
I don't know what to do next. Thank You. 




 ------ Converting 9 and 27 to base 3 





x + 2x - 2 = 3 - 3x ------ Bases are equal and so are their EXPONENTS

3x + 3x = 3 + 2

6x = 5

 

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