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froilan041
Sep21-08, 06:05 PM
Can anyone show me how to solve the following equations:

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

3^(x+1)+3^(x)=36
I would greatly appreciate your assistance.

rock.freak667
Sep21-08, 06:11 PM
2x-1=2x*2-1.

It should be a bit easier now.

SnipedYou
Sep21-08, 09:39 PM
Here is a general way to solve things like this
1. 2^{x-1}-2^{x} = 2^{-3}
Now factor out a 2^{x-1}
2^{x-1}(1 - 2) = 2^{3} \rightarrow 2^{x - 1} = -\frac{1}{8}
You can take the logarithm at this point or you could notice that -\frac{1}{8}
is out of the range of 2^{x}

Similar approach
2. 3^{x+1} + 3^{x} = 36 \rightarrow
3^{x}(3 + 1) = 36 \rightarrow
3^{x} = 9 \rightarrow
x = 2