SOLUTION: Can you please help me figure out how to solve this problem? log(40x+200)=3 Here is what I came up with. log40x+log200=3 log40x+2.301=3 log40x=.699 x=.4363 When I plug X ba

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Can you please help me figure out how to solve this problem? log(40x+200)=3 Here is what I came up with. log40x+log200=3 log40x+2.301=3 log40x=.699 x=.4363 When I plug X ba      Log On


   

Question 526762: Can you please help me figure out how to solve this problem?
log(40x+200)=3
Here is what I came up with.
log40x+log200=3
log40x+2.301=3
log40x=.699
x=.4363
When I plug X back into the equation, it doesn't come out right.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
It's not surprising that it doesn't come out right!
Because:
Log%2840x%2B200%29is not equal to++Log%2840x%29%2BLog%28200%29
Here's what you must do:
Log%2840x%2B200%29+=+3 Change this to the exponential form:If Log%5Bb%5D%28y%29+=+x then y+=+b%5Ex
Log%2840x%2B200%29+=+3---->%2840x%2B200%29+=+10%5E3 Simplify:
40x%2B200+=+1000 Subtract 200 from both sides.
40x+=+800 Divide both sides by 40.
x+=+20