document.write( "Question 243826: how many real solutions does the following equation have?\r
\n" ); document.write( "\n" ); document.write( "log(4x-15)=logx+log(x+4)
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Algebra.Com's Answer #178598 by Theo(13342)\"\" \"About 
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log(4x-15) = log(x) + log(x+4)\r
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\n" ); document.write( "\n" ); document.write( "log(4x-15) = log(x*(x+4)) because log(a*b) = log(a) + log(b)\r
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\n" ); document.write( "\n" ); document.write( "In this case, a was equal to x and b was equal to (x+4)\r
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\n" ); document.write( "\n" ); document.write( "if log(4x-15) = log(x*(x+4)) then:\r
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\n" ); document.write( "\n" ); document.write( "4x-15 = x*(x+4) which becomes:\r
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\n" ); document.write( "\n" ); document.write( "4x-15 = x^2 + 4x\r
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\n" ); document.write( "\n" ); document.write( "subtract 4x from both sides of this equation to get:\r
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\n" ); document.write( "\n" ); document.write( "x^2 + 4x - 4x = -15 which becomes:\r
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\n" ); document.write( "\n" ); document.write( "x^2 = -15\r
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\n" ); document.write( "\n" ); document.write( "take the square root of both sides of this equation to get:\r
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\n" ); document.write( "\n" ); document.write( "x = +/- sqrt(-15)\r
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\n" ); document.write( "\n" ); document.write( "since you are taking the square root of a negative number, and the result of that is not real, then the number of real solutions that this equation has is 0.
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