SOLUTION: LogBase4(x-2)+LogBase4(x+10)=3

Question 730965: LogBase4(x-2)+LogBase4(x+10)=3
Answer by fcabanski(1391) (Show Source): You can put this solution on YOUR website!
Remember that log(base n) (x) + log(base n) (y) = log(base n) (xy)

LogBase4(x-2)+LogBase4(x+10)=3

logbase((x-2)*(x+10)) = 3

logbase4(x^2 + 8x -20) = 3

Recall that logbasen (x) = y if and only if n^y = x.

(x+14)(x-6) = 0

x=-14 and x=6

The log of a negative number is imaginary. So only x=6 works. -14 results in logbase4 (-16) + logbase4 (-4) = 3 . The log of negative numbers isn't real.

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