Solving for 2-2^-x in 1/2^x+(2^(x)-1)/2^(x-1)

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The discussion revolves around solving the equation 1/2^x + (2^x - 1)/2^(x-1) to equal 2 - 2^-x. The initial approach involved using index laws to simplify the expression to 2^-x + 2^(1-x)(2^x - 1). After expanding, the result was 2^-x + 2 - 2^(1-x), but further steps were unclear. Participants suggest factoring a 2 out of 2^(1-x) to aid in solving the equation. The focus remains on simplifying and rearranging terms to achieve the desired equality.
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Homework Statement


I need to get 1/2^x+(2^(x)-1)/2^(x-1) to equal 2-2^-x
I originally used index laws so 2^-x+2^(1-x)[2^x - 1]
From there i expanded so that 2^-x + 2 - 2^(1-x) was the result, I'm not sure where to go from here
 
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try bringing the 2^-x on the RHS to the LHS and then proceed. No need to use index laws here till the very end.
 
jackscholar said:

Homework Statement


I need to get 1/2^x+(2^(x)-1)/2^(x-1) to equal 2-2^-x
I originally used index laws so 2^-x+2^(1-x)[2^x - 1]
From there i expanded so that 2^-x + 2 - 2^(1-x) was the result, I'm not sure where to go from here
I assume you want to show that \displaystyle \ \ 2^{-x}+\frac{2^{x}-1}{2^{x-1}}=2-2^{-x}\ .

Your result looks correct so far.

Factor a 2 out of \displaystyle \ \ 2^{1-x}\ .
 
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