SOLUTION: im not sure of how to solve these problems, 1. log3(2x-1) + log(base 3)2 = 2 2. 4(^3-2x = 12 3. 3^x = 1/81 4. 9^x - 6x3^x - 27 = 0 all working would be greatly appreciated

Question 460753: im not sure of how to solve these problems,
1. log3(2x-1) + log(base 3)2 = 2
2. 4(^3-2x = 12
3. 3^x = 1/81
4. 9^x - 6x3^x - 27 = 0
all working would be greatly appreciated,
emily.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
1. log3(2x-1) + log(base 3)2 = 2
2. 4(^3-2x = 12
3. 3^x = 1/81
4. 9^x - 6x3^x - 27 = 0
..
1. log3(2x-1) + log(base 3)2 = 2
Write as single log
log3(2x-1)(2)=2
convert to exponential form:(base(3) raised to log of number(2)=number((2x-1)(2)
3^2=(2x-1)(2)
9=4x-2
4x=11
x=11/4
..
2. 4(^3-2x = 12
4^3-2x=12
64-2x=12
2x=64-12=52
x=26
..
3. 3^x = 1/81
3^x=1/3^4=3^-4
x=-4
..
4. 9^x - 6x3^x - 27 = 0
I believe this equation was meant to be written as follows:
9^x - (6)3^x - 27 = 0
3^2x-6*3^x-27=0
(3^x+3)(3^x-9)=0
3^x+3=0
3^x≠-3 (reject)
3^x-9=0
3^x=9=3^2
x=2
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