document.write( "Question 35832: Solve: ln(3-y)-ln(2y+1)=ln(4y) \n" ); document.write( "

Algebra.Com's Answer #21930 by narayaba(40) $\"\"$ $\"About$

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ln(3-y)-ln(2y+1)=ln(4y)\r
\n" ); document.write( "\n" ); document.write( "ln(a) - ln(b) can be written as ln(a/b)
\n" ); document.write( "similarly
\n" ); document.write( "ln(3-y) - ln(2y+1) = ln((3-y)/(2y+1)) = ln(4y)\r
\n" ); document.write( "\n" ); document.write( "ln((3-y)/(2y+1)) = ln(4y)\r
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\n" ); document.write( "\n" ); document.write( "logrrathims on both aides are to the same base
\n" ); document.write( "therefore
\n" ); document.write( "(3-y) / (2y+1) = 4y\r
\n" ); document.write( "\n" ); document.write( "3-y = (2y+1)*(4y)\r
\n" ); document.write( "\n" ); document.write( "3-y = 8y^2 + 4y\r
\n" ); document.write( "\n" ); document.write( "8y^2 + 5y -3 = 0 (you can write 5y as 8y - 3y, just makes it easy to solve it)
\n" ); document.write( "8y^2 +8y - 3y -3 = 0
\n" ); document.write( "(8y-3)(y+1) = 0
\n" ); document.write( "y = 3/8 or -1\r
\n" ); document.write( "\n" ); document.write( "to solve 8y^2 + 5y -3 = 0
\n" ); document.write( "you can also use the formula $\"y+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29++\"$
\n" ); document.write( "and find y
\n" ); document.write( "b = 5, a = 8 and c = -3\r
\n" ); document.write( "\n" ); document.write( "ln of negative nunber is not possible so y = 3/8\r
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