SOLUTION: Logx-log16=log2x + logx Solve for x

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Logx-log16=log2x + logx Solve for x       Log On


   

Question 1070581: Logx-log16=log2x + logx
Solve for x
Found 2 solutions by josmiceli, Boreal:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+log%28+x+%29+-+log%28+16+%29+=+log%28+2x+%29+%2B+log%28+x+%29+
Subtract +log%28x%29+ from both sides
+-log%28%28+16+%29%29+=+log%28%28+2x+%29%29+
+log%28%28+16%5E%28-1%29+%29%29+=+log%28%28+2x+%29%29+
+log%28%28+1%2F16+%29%29+=+log%28%28+2x+%29%29+
+1%2F16+=+2x+
+x+=+1%2F32+
-------------------
check answer:
+log%28+x+%29+-+log%28+16+%29+=+log%28+2x+%29+%2B+log%28+x+%29+
+log%28+.03125+%29+-+log%28+16+%29+=+log%28+.0625+%29+%2B+log%28+.03125+%29+
+-1.50515+-+1.20412+=+-1.20412+-+1.50515+
+-2.70927+=+-2.70927+
OK

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
that is log (x/16)= log (2x^2)
2x^2=x/16
x=1/32 ANSWER
log (1/32)-log 16=-2.709
log (1/512)=-2.709