document.write( "Question 254819: How do I go about this question? \r
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\n" ); document.write( "\n" ); document.write( "log2(x2 + 4x + 3) – log 2(x2 + x) = 4
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Algebra.Com's Answer #187118 by Theo(13342)\"\" \"About 
You can put this solution on YOUR website!
problem is:\r
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\n" ); document.write( "\n" ); document.write( "\"log%282%2Cx%5E2+%2B+4x+%2B+3%29+-+log%282%2Cx%5E2+%2B+x%29+=+4\"\r
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\n" ); document.write( "\n" ); document.write( "since \"log%28a%2C%28b%2Fc%29%29+=+log%28a%2Cb%29+-+log%28a%2Cc%29\", then the reverse is also true, i.e.:\r
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\n" ); document.write( "\n" ); document.write( "\"log%28a%2Cb%29+-+log%28a%2Cc%29+=+log%28a%2C%28b%2Fc%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "applying this rule of logarithms makes your equation become:\r
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\n" ); document.write( "\n" ); document.write( "\"log%282%2C%28x%5E2+%2B+4x+%2B+3%29%2F%28x%5E2+%2B+x%29%29+=+4\"\r
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\n" ); document.write( "\n" ); document.write( "since \"y+=+log%282%2Cx%29\" if and only if \"2%5Ey+=+x\", applying this rule makes your equation become:\r
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\n" ); document.write( "\n" ); document.write( "\"2%5E4+=+%28x%5E2+%2B+4x+%2B+3%29%2F%28x%5E2+%2B+x%29\"\r
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\n" ); document.write( "\n" ); document.write( "multiply both sides of this equation by \"x%5E2+%2B+x\" and you get:\r
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\n" ); document.write( "\n" ); document.write( "\"2%5E4+%2A+%28x%5E2+%2B+x%29=+%28x%5E2+%2B+4x+%2B+3%29\"\r
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\n" ); document.write( "\n" ); document.write( "multiply the factors out and you get:\r
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\n" ); document.write( "\n" ); document.write( "\"16%2Ax%5E2+%2B+16%2Ax+=+x%5E2+%2B+4%2Ax+%2B+3\"\r
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\n" ); document.write( "\n" ); document.write( "subtract \"x%5E2+%2B+4%2Ax+%2B+3\" from both sides of this equation and combine like terms to get:\r
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\n" ); document.write( "\n" ); document.write( "\"15%2Ax%5E2+%2B+12%2Ax+-+3+=+0\"\r
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\n" ); document.write( "\n" ); document.write( "factor this equation to get:\r
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\n" ); document.write( "\n" ); document.write( "\"15%2Ax+-+3%29+%2A+%28x+%2B+1%29+=+0\"\r
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\n" ); document.write( "\n" ); document.write( "solve for x to get:\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%281%2F5%29\" or \"x+=+-1\"\r
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\n" ); document.write( "\n" ); document.write( "confirm by substituting in original equation of:\r
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\n" ); document.write( "\n" ); document.write( "\"log%282%2Cx%5E2+%2B+4x+%2B+3%29+-+log%282%2Cx%5E2+%2B+x%29+=+4\"\r
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\n" ); document.write( "\n" ); document.write( "to confirm using your calculator, you need to convert the base of the logarithm from 2 to 10 or e.\r
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\n" ); document.write( "\n" ); document.write( "I used 10 because it is the LOG function of my calculator.\r
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\n" ); document.write( "\n" ); document.write( "the conversion formula says:\r
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\n" ); document.write( "\n" ); document.write( "\"log%28a%2Cb%29+=+log%28c%2Cb%29%2Flog%28c%2Ca%29\"\r
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\n" ); document.write( "\n" ); document.write( "if we let a = base of 2 and we let c = base of 10, this formula becomes:\r
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\n" ); document.write( "\n" ); document.write( "\"log%282%2Cb%29+=+log%2810%2Cb%29%2Flog%2810%2C2%29\"\r
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\n" ); document.write( "\n" ); document.write( "your formula of \"log%282%2Cx%5E2+%2B+4x+%2B+3%29+-+log%282%2Cx%5E2+%2B+x%29+=+4\" becomes:\r
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\n" ); document.write( "\n" ); document.write( "substituting x = (1/5) into this equation, we get:\r
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\n" ); document.write( "\n" ); document.write( "this becomes:\r
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\n" ); document.write( "\n" ); document.write( "\"1.941106311+-+%28-2.058893689%29+=+4\" which becomes:\r
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\n" ); document.write( "\n" ); document.write( "\"4+=+4\" confirming that x = (1/5) is good.\r
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\n" ); document.write( "\n" ); document.write( "when x = -1, confirmation was not successful because:\r
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\n" ); document.write( "\n" ); document.write( "original equation is:\r
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\n" ); document.write( "\n" ); document.write( "\"log%282%2Cx%5E2+%2B+4x+%2B+3%29+-+log%282%2Cx%5E2+%2B+x%29+=+4\"\r
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\n" ); document.write( "\n" ); document.write( "substitute -1 for x to get:\r
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\n" ); document.write( "\n" ); document.write( "this becomes:\r
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\n" ); document.write( "\n" ); document.write( "\"log%282%2C0%29+-+log%282%2C0%29+=+4\"\r
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\n" ); document.write( "\n" ); document.write( "since you can't take log of 0 to any base, x = -1 is not a valid answer.\r
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\n" ); document.write( "\n" ); document.write( "graph of your original equation looks like this:\r
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\n" ); document.write( "\n" ); document.write( "the answer to your problem is:\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%281%2F5%29\"\r
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