SOLUTION: solve log(3x+4)=2log(x)

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Question 507520: solve log(3x+4)=2log(x)
Found 2 solutions by nerdybill, Earlsdon:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
log(3x+4)=2log(x)
log(3x+4)=log(x^2)
3x+4 = x^2
0 = x^2 - 3x - 4
0 = (x-4)(x+1)
x = {4, -1}
we can throw out the -1 solution (extraneous) leaving:
x = 4

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve:
Log%283x%2B4%29+=+2Log%28x%29 Apply the "power rule" to the right side.
Log%283x%2B4%29+=+Log%28x%5E2%29 Now if Log%28M%29+=+Log%28N%29 then M+=+N so...
3x%2B4+=+x%5E2 Rearrange as:
x%5E2-3x-4+=+0 Solve by factoring:
%28x%2B1%29%28x-4%29+=+0 Apply the "zero product" rule:
x%2B1+=+0 or x-4+=+0 so that...
x+=+-1 or x+=+4
Now you must check the results for "extraneous" or "invalid" values:
Try x+=+-1
Log%283x%2B4%29+=+2Log%28x%29 Substitute x+=+-1
Log%283%28-1%29%2B4%29+=+2Log%28-1%29
Log%281%29+=+2Log%28-1%29
0+=+2%281.364376%29i This is not valid!
Try x+=+4
Log%283x%2B4%29+=+2Log%28x%29 Substitute x+=+4
Log%2816%29+=+2Log%284%29
1.20411998+=+2%280.60205999%29
1.20411998+=+1.20411998 This is valid.
Solution is x+=+4