SOLUTION: log base3 x +log base3 (x-4) +16=18

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Question 21665: log base3 x +log base3 (x-4) +16=18
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x:
log3(x)+log3(x-4)+16 = 18 Subtract 16 from both sides.
log3(x)+log3(x-4) = 2 Apply the product rule for logarithms:LogbM+logbN = logb(MN)
log3(x)+log3(x-4) = log3(x^2-4x) = 2 Rewrite in exponential form:
logbx = y means b%5Ey+=+x so:
log3(x^2-4x) = 2 means 3%5E2+=+%28x%5E2-4x%29 Subtract 3^2 = 9 from both sides.
x%5E2-4x-9+=+0 Solve using the quadratic formula: x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a
x+=+%28-%28-4%29%2B-sqrt%28%28-4%29%5E2-4%281%29%28-9%29%29%29%2F2%281%29
x+=+%284%2B-sqrt%2852%29%29%2F2 = %284%2B-sqrt%284%2A13%29%29%2F2
x+=+2%2B-sqrt%2813%29
The roots are:
x+=+2%2Bsqrt%2813%29
x+=+2-sqrt%2813%29