SOLUTION: Solve for x: log(4x^2)-log(16x)=1

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Question 49208: Solve for x: log(4x^2)-log(16x)=1

Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
+log%284x%5E2%29-log%2816x%29=1+
+log%28%284x%5E2%29%2F%2816x%29%29+=+1+

Assuming the log is base 10, then to remove the log we next to raise both sides to the base 10:
+%284x%5E2%29%2F%2816x%29+=+10%5E1+
+%284x%5E2%29%2F%2816x%29+=+10+

this is where the algebraic method and logs seem to me to be at odds to each other. Looking at the above line, we could do one of 2 approaches. Either divide the lefthand side first or move the 16x up to the righthand side.

Approach1:
+%284x%5E2%29%2F%2816x%29+=+10+
+%28x%29%2F%284%29+=+10+
+x+=+40+

Approach2:
+%28x%5E2%29%2F%284x%29+=+10+
+x%5E2+=+40x+
+x%5E2+-+40x+=+0+
+x%28x+-+40%29+=+0+
so either x=0 or x-40=0
--> x=0 or x=40

This second approach seems to give us 2 answers whereas the first gave us just 1. However, putting x=0 into the original will give you an error, so the only solution is x=40... you must remember to check by substituting your answers back into the question, just to check.

jon.