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Solve 3^4x − 1 = 9^x + 1?
How do I do this? Can someone explain the process in steps. When I tried to solve it I got 2/3; it was wrong
I meant 3^(4x-1) = 9^(x+1), sorry for not writing it correct. How did you get 3/2?
1 Answer
- KevinMLv 79 years agoFavorite Answer
Ok - first of all, I want to be clear about the question: Do you mean 3^(4x) - 1 = 9^(x) + 1? If so, then the answer is:
9^(2x) - 9^(x) - 2 = 0; u = 9^x
u^2 - u - 2 = 0
(u-2)(u+1) = 0
u = -1 or 2, but only 2 works, so:
9^x = 2
x = log_9 (2) = log(2) / log(9)
If you actually meant 3^(4x-1) = 9^(x+1), then x = 3/2 (not 2/3).
Ed: 3^(4x-1) = 3^2(x+1) = 3^(2x+2)
4x-1 = 2x+2
2x = 3
x = 3/2
