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2^y x 8^-y = 4^y
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1. When you see exponents with different bases, make them the same base. (Notice that 4=2^2 and 8=2^3)
Do that first, what do you get? :)
2^y x (2^3)^-y = (2^2)^y
Now, use the law of indices: \(\Large (a^b)^n=a^{bn}\)
2^y x 2^-3y = 2^2y would that be 2^-2y = 2^2y ?
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yes, you're getting it so fast :)
Now use this law of indices: \(\Large a^{-n}=\frac1{a^n}\)
\[\frac{ 1 }{ 2^{2y} } = 2^{2y}\]
Now leave the 1 in the LHS and move the other stuffs to the RHS :)
\[1 = 2^{4x ^{2}}\]
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Where did you get the x^2 lol
i meant y. sorry
lolz yep, now notice that \(1=2^0\)
ohhhh
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