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\[(y^3)^4 = y ^{3*4}\]
\[x ^{2-1}y ^{12-5}\]
yes
ok, so?? compare to Kainui, I am nothing. hihihihi
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oh well hi u oops & now another one @Kainui \[\frac{ 2^5⋅8^4}{16}=\frac{ 2^5⋅(2^a)^4}{ 2^4 }= \frac{ 2^5⋅2^b }{ 2^4 }= 2^c\]
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