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If x > 0, then (4^x)(8^x) = ?
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If x > 0 then, \[(4^{x}) (8^{x})\] =
\[4=2^2\] \[8=2^3\] \[(2^2)^x*(2^3)^x\]
(2^2x)*(2^3x)=2^(2x+3x)=2^(5x)
@SlightlyConfused
thanks! still a bit confused though but thanks
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tell me buddy @SlightlyConfused
I'm confused on this part; (2^2x)*(2^3x)=2^(2x+3x)=2^(5x)
I got the first part, but how did you go from there to that long equation?
never mind i got it! you multiplied the 2 and the x!!!!!!!!!!!
if (2^a)^b then powers will be multiplied (2^a)^b=2^(a*b)
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now if base is same then power will be added we converted it to a common base 2 so thsat we can add up the powers
@SlightlyConfused ?????
okay i got it! thanks so much!!!!!!!!!!
alwys welcome :)
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