SOLUTION: solve (3^2x) -[5(3^x)]=-6 I've tried a few different roots..nothing seems to be working out :? this is what i've tried log (3^2x -5(3^x)+6) =0 2xlog3-log5(3^x)+log6 =0

Question 868808: solve (3^2x) -[5(3^x)]=-6
I've tried a few different roots..nothing seems to be working out :?
this is what i've tried
log (3^2x -5(3^x)+6) =0
2xlog3-log5(3^x)+log6 =0
2xlog3-(log5+log3^x) =-log6
2xlog3 -log5 -xlog3 = -log6
x(2log3-log3) =-log6+log5
x =(-log6+log5)/(log3)
the answer is 1 and it works...
I tried squaring the whole equation cause there's another answer of 0.631
I've tried various ways...nothing's getting me anywhere.
a solution would be greatly appreciated. I don't need this for an assignment or test just for curiosity.
Thanks

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
solve (3^2x) -[5(3^x)]=-6
(3^2x) + 6 = 5*3^x
------------
You can't say
log(3^2x) + log(6) = log(5*3^x)
The log of a sum is not the sum of the logs.
That's where you went wrong.
------------
(3^2x) -[5(3^x)]=-6
(3^2x) - 5(3^x) + 6 = 0
That cannot be solved by algebraic methods.
You can use a graphing calculator or iterative methods.
1 is a zero.
0.63092975357146 is the second zero.

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