document.write( "Question 80522: Solve for x: log(base 5)3x + log(base 5)(x-3) = 1\r
\n" ); document.write( "\n" ); document.write( "I see they both have a common base of 5. I'm not sure if I'm on the right track, but this is what I got so far:
\n" ); document.write( "log(base 5)[3x(x-3)] = 1
\n" ); document.write( "3x(x-3) = 5^1
\n" ); document.write( "3x^2 - 9x =5
\n" ); document.write( "3x^2 - 9x - 5 = 0
\n" ); document.write( "I then tried factoring it out, but I don't think it works.\r
\n" ); document.write( "\n" ); document.write( "I think that first 3 in the problem might be throwing me off for some reason. Any help with this is greatly appreciated. Thanks.
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Algebra.Com's Answer #57756 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
Solve for x:
\n" ); document.write( "\"Log%5B5%5D%283x%29%2BLog%5B5%5D%28x-3%29+=+1\" Apply the product rule for logarithms.
\n" ); document.write( "\"Log%5B5%5D%28%283x%29%28x-3%29%29+=+1\" Simplify the left side.
\n" ); document.write( "\"Log%5B5%5D%283x%5E2-9x%29+=+1\" Rewrite in exponential form.
\n" ); document.write( "\"5%5E1+=+3x%5E2-9x\" Subtract 5 from both sides.
\n" ); document.write( "\"3x%5E2-9x-5+=+0\" Your work up to this point is commendable!...and you are correct, this quadratic equation is not factorable, so you can use the quadratic formula: \"x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a\"
\n" ); document.write( "\"x+=+%28-%28-9%29%2B-sqrt%28%28-9%29%5E2-4%283%29%28-5%29%29%29%2F2%283%29\"
\n" ); document.write( "\"x+=+%289%2B-sqrt%2881%2B60%29%29%2F6\"
\n" ); document.write( "\"x+=+%289%2B-sqrt%28141%29%29%2F6\"
\n" ); document.write( "The exact answers are:
\n" ); document.write( "\"x+=+3%2F2%2B%28sqrt%28141%29%29%2F6\"
\n" ); document.write( "\"x+=+3%2F2-%28sqrt%28141%29%29%2F6\"
\n" ); document.write( "The approximate answers are:
\n" ); document.write( "\"x+=+3.48\" To the nearest hundredth.
\n" ); document.write( "\"x+=+-0.48\" To the nearest hundredth.
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