document.write( "Question 526762: Can you please help me figure out how to solve this problem?
\n" ); document.write( "log(40x+200)=3
\n" ); document.write( "Here is what I came up with.
\n" ); document.write( "log40x+log200=3
\n" ); document.write( "log40x+2.301=3
\n" ); document.write( "log40x=.699
\n" ); document.write( "x=.4363
\n" ); document.write( "When I plug X back into the equation, it doesn't come out right.
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Algebra.Com's Answer #348700 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
It's not surprising that it doesn't come out right!
\n" ); document.write( "Because:
\n" ); document.write( "\"Log%2840x%2B200%29\"is not equal to\"++Log%2840x%29%2BLog%28200%29\"
\n" ); document.write( "Here's what you must do:
\n" ); document.write( "\"Log%2840x%2B200%29+=+3\" Change this to the exponential form:If \"Log%5Bb%5D%28y%29+=+x\" then \"y+=+b%5Ex\"
\n" ); document.write( "\"Log%2840x%2B200%29+=+3\"---->\"%2840x%2B200%29+=+10%5E3\" Simplify:
\n" ); document.write( "\"40x%2B200+=+1000\" Subtract 200 from both sides.
\n" ); document.write( "\"40x+=+800\" Divide both sides by 40.
\n" ); document.write( "\"x+=+20\"
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