document.write( "Question 389428: log2(2x+3)+log2x=1 \n" ); document.write( "
Algebra.Com's Answer #275956 by CharlesG2(834)\"\" \"About 
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log2(2x+3)+log2x=1\r
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\n" ); document.write( "\n" ); document.write( "assuming base 2 on the logs
\n" ); document.write( "log2 (2x + 3) + log2 (x) = 1
\n" ); document.write( "logarithmic rule: logb (m) + logb (n) = logb (mn)
\n" ); document.write( "log2 x(2x + 3) = 1
\n" ); document.write( "logarithmic rule: if logb (y) = x then b^x = y
\n" ); document.write( "2^1 = 2 = x(2x + 3) = 2x^2 + 3x
\n" ); document.write( "0 = 2x^2 + 3x - 2
\n" ); document.write( "0 = (2x - 1)(x + 2) by FOIL --> 2x^2 - x + 4x - 2, same as above
\n" ); document.write( "2x - 1 = 0 --> 2x = 1 --> x = 1/2 or x + 2 = 0 --> x = -2
\n" ); document.write( "check x = 1/2:
\n" ); document.write( "log2 (2 * 1/2 + 3) + log2 (1/2) = 1
\n" ); document.write( "log2 (1 + 3) + log2 (1/2) = 1
\n" ); document.write( "log2 (4) + log2 (1/2) = log2 (2) = 1, yes
\n" ); document.write( "check x = -2:
\n" ); document.write( "log2 (2 * -2 + 3) + log2 (-2) = 1
\n" ); document.write( "log2 (-4 + 3) + log2 (-2) = 1
\n" ); document.write( "log2 (-1) + log2 (-2) = 1, no can not take logs of negative numbers
\n" ); document.write( "solution is only x = 1/2\r
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