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if 8^x+8^x-1=10 then 2^2x
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2.8^x=11 8^x=11/2take log x=(log11/2)/log 8 I think so but it looks like an improper method :)
or is the equation like \[8^{x}+8^{x-1}=10\]
if 8x+8x−1=10 then 2^2x...
What happened to the ^ sign in the question?
^ rank
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Anyway if that is the question then 16x=11 x=11/16 2^2x=2^(11/8)
2 \[8^{x}+8^{x-1}=10 then 2^{2x}\]
ah ok so \[8^{x}+8^{x}/8=10\]so 9.8^x=80 8^x=80/9 2^3x=80/9, powering both sides bt 2/3 2^2x=(80/9)^2/3
ok..thanks :)
no problem :)
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