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solve: 2^x=8^x+2
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2^x = (2^3)^x + 2
2^x = 2^(3x) + 2
factor out 2
2 (2^(x-1) ) = 2 (2^(3x-1) ) + 2
cancel off 2s
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2^(x-1) = 2^(3x-1) +1
I would probably move both over to the LHS
2^(x-1) -2^(3x-1) = 1
take ln of both sides
ln [ 2^(x-1) -2^(3x-1) ] = ln(1)
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dont think that did anything :|
This has no solutions
small proof would be let a = 2^x, so then a^3 = 8^x, so we get a^2 = a^3 + 2 so a^3 - a^2 + 2 = 0, from judge if a = -1, then -1 -1 + 2 = 0, so a = -1, then 2^x = -1, which is impossible, since 2^x is always more then 0. hope this helps, favourite me if it did :)
You there, does that make sense?
Yea, i think i get it, working it out again to make sure!! thank you both.
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