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If 4^(2-x)=3 then 4^(2x-1)=?
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-x + 2 = log_4(3) 2x - 1 = log_4(?) x + 1 = log_4(3*?) 3*? = 4^(x+1) ? = (4^(x+1))/3
should be a way to cancel x 1 minute
-2x + 4 = log_4(9) 2x - 1 = log_4(y) 3 = log_4(9y) 9y = 64 y = 64/9
how do you know where to add the log?
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which part?
pls explain this part futher: -2x + 4 = log_4(9) 2x - 1 = log_4(y) 3 = log_4(9y)
(4^(2-x))^2=3^2 4^(4-2x)=9 4^4=9*4^2x 4^3=9*4^(2x-1) 4^(2x-1)=64/9
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