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log2(5x^2-8x)=2 find x
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question...are you allowed to use a calculator or no?
no
anybody know?
apply base 2 exponential to both sides, it cancels the log 5x^2-8x=2^2
so square both sides? i don't understand?
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hang on it looks like this right \[\log _{2}(5x^{2}-8x)=2\]
yes
Alright, your log base is 2, to undo that you do this \[\large 2^{\log_{2}(5x^{2}-8x)}=2^2\]
the left 2 and log2 cancel out and leaves 5x^2-8x=4
x= 12/5?
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No, move the 4 over 5x^2-8x-4=0 it factors as (x-2)(5x+2) so solve x-2=0 and 5x+2=0
x = 2 , x = -2/5
Right
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