SOLUTION: can somebody help me solve this equation {{{2^(x+1)8^(-x) =4}}}

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Question 169003This question is from textbook college algebra
: can somebody help me solve this equation 2%5E%28x%2B1%298%5E%28-x%29+=4This question is from textbook college algebra

Found 2 solutions by Alan3354, Earlsdon:
Answer by Alan3354(21611) About Me  (Show Source):
You can put this solution on YOUR website!
2%5E%28x%2B1%298%5E%28-x%29+=4
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2%5E%28x%2B1%292%5E%28-3x%29+=2%5E2
2%5E%281-2x%29+=2%5E2
1-2x = 2
x = -1/2

Answer by Earlsdon(6103) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x:
2%5E%28x%2B1%29%2A8%5E%28-x%29+=+4 Substitute 8+=+2%5E3 and 4+=+2%5E2
2%5E%28x%2B1%29%2A2%5E%28-3x%29+=+2%5E2 Perform the indicated multiplication by adding the exponents.
2%5E%28x%2B1-3x%29+=+2%5E2 Simplify the left side.
2%5E%281-2x%29+=+2%5E2 The bases (2) are equal so the exponents are equal.
1-2x+=+2 Subtract 1 from both sides.
-2x+=+1 Divide both sides by -2.
highlight%28x+=+-1%2F2%29
Check:
2%5E%28x%2B1%29%2A8%5E%28-x%29+=+4 Substitute x+=+-1%2F2
2%5E%28-%281%2F2%29%2B1%29%2A8%5E%28-%28-1%2F2%29%29+=+4 Substitute 8+=+2%5E3 and simplify the left side.
2%5E%281%2F2%29%2A2%5E%283%2F2%29+=+4 Perform the indicated multiplication by adding the exponents.
2%5E%284%2F2%29+=+4
2%5E2+=+4
4+=+4