document.write( "Question 1203539: Suppose that x=ln(A) and y=ln(B). Write the following formula in terms of x and y \r
\n" ); document.write( "\n" ); document.write( "ln(A-B)=?\r
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\n" ); document.write( "\n" ); document.write( "Hi everyone, I have been struggling with trying to figure out how to solve this. Any help and explaining how you solved it would be helpful :)
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Algebra.Com's Answer #839185 by math_tutor2020(3817)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "One log rule is ln(A)-ln(B) = ln(A/B)\r
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\n" ); document.write( "\n" ); document.write( "But there isn't a log rule for ln(A-B)\r
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\n" ); document.write( "\n" ); document.write( "I think there might be a typo in your textbook.\r
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\n" ); document.write( "\n" ); document.write( "I suppose this could be one pathway to take, but it's a bit convoluted and messy.
\n" ); document.write( "x = ln(A)
\n" ); document.write( "y = ln(B)
\n" ); document.write( "A = e^x
\n" ); document.write( "B = e^y
\n" ); document.write( "A-B = e^x - e^y
\n" ); document.write( "A-B = e^x(1 - e^(y-x))
\n" ); document.write( "Ln[A-B] = Ln[ e^x(1 - e^(y-x)) ]
\n" ); document.write( "Ln[A-B] = Ln[e^x] + Ln[1 - e^(y-x) ]
\n" ); document.write( "Ln[A-B] = 1 + Ln[1 - e^(y-x)]
\n" ); document.write( "But we run into the same problem of having a log of the form Ln(something - somethingElse)
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