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solve for x \[8^{x-1} = 2^{x+1}\]
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replace 8 with 2^3 and go from there...
0.o?
\[2^{3^{x-1}} = 2^{x+1}\] ?
rule of exponents i mean
Pretty close. 3*(x-1) not 3x-1
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it's not 3^x-1?
\[(2^{3})^{x-1}= 2^{3(x-1)}\]
wait I seem to remember if the bases are the same ( both 2) then it should just be 3(x-1) = x+1 right?
yes.
x=2?
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Yes. But you can (and should) check it. plug 2 into the original problem
gives me 8 = 2 :(
oh wait
hey it's right!
thanks phi!
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