document.write( "Question 173045: Please explain how to work this problem:
\n" ); document.write( "log of(x+1), base 2 plus 2 times the log of(x-1), base 4 equals 3
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Algebra.Com's Answer #127901 by gonzo(654)\"\" \"About 
You can put this solution on YOUR website!
i think i have a solution for you.
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\n" ); document.write( "before i go any further, your answer is:
\n" ); document.write( "x = 3
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\n" ); document.write( "i checked that out in the original formulas and it works.
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\n" ); document.write( "the problem:
\n" ); document.write( "\"log%282%2C%28x%2B1%29%29+%2B+2%2Alog%284%2C%28x-1%29%29+=+3\"
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\n" ); document.write( "so i could see the problem a little easier, i let:
\n" ); document.write( "a = (x+1)
\n" ); document.write( "and
\n" ); document.write( "b = (x-1)
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\n" ); document.write( "the problem then looks like:
\n" ); document.write( "\"log%282%2Ca%29+%2B+2%2Alog%284%2Cb%29+=+3\"
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\n" ); document.write( "i then wanted to get a common base, so i converted the logarithm with a base of 4 to a logarithm with a base of 2 using the conversion formula:
\n" ); document.write( "\"log%284%2Cb%29+=+log%282%2Cb%29%2Flog%282%2C4%29\"
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\n" ); document.write( "the problem then looks like:
\n" ); document.write( "\"log%282%2Ca%29+%2B+2+%2A+%28log%282%2Cb%29%2Flog%282%2C4%29%29+=+3\"
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\n" ); document.write( "since \"log%282%2C4%29+=+d\" if and only if \"2%5Ed+=+4\" and i can show that d has to be equal to 2, then:
\n" ); document.write( "\"log%282%2C4%29+=+2\"
\n" ); document.write( "and the problem now looks like:
\n" ); document.write( "\"log%282%2Ca%29+%2B+2%2A%28log%282%2Cb%29%2F2%29+=+3\"
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\n" ); document.write( "this simplifies to:
\n" ); document.write( "\"log%282%2Ca%29+%2B+log%282%2Cb%29+=+3\"
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\n" ); document.write( "since the laws of logarithms state:
\n" ); document.write( "\"log%282%2Ca%2Ab%29+=+log%282%2Ca%29+%2B+log%282%2Cb%29\"
\n" ); document.write( "then:
\n" ); document.write( "\"log%282%2Ca%29+%2B+log%282%2Cb%29\" must equal \"log%282%2Ca%2Ab%29\"
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\n" ); document.write( "the problem now looks like:
\n" ); document.write( "\"log%282%2Ca%2Ab%29+=+3\"
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\n" ); document.write( "the basic laws of logarithms state:
\n" ); document.write( "\"log%282%2Ca%2Ab%29+=+3\" if and only if:
\n" ); document.write( "\"2%5E3+=+a%2Ab\"
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\n" ); document.write( "since \"2%5E3\" = 8, the problem now looks like:
\n" ); document.write( "\"a%2Ab+=+8\"
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\n" ); document.write( "if you recall from way back, i let:
\n" ); document.write( "a = (x+1)
\n" ); document.write( "b = (x-1)
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\n" ); document.write( "i can now substitute (x+1) for a, and (x-1) for b, and the problem now looks like:
\n" ); document.write( "\"%28x%2B1%29%2A%28x-1%29+=+8\"
\n" ); document.write( "multiplying out the factors, this becomes:
\n" ); document.write( "\"x%5E2+-1+=+8\"
\n" ); document.write( "add 1 to both sides of the equation to get:
\n" ); document.write( "\"x%5E2+=+9\"
\n" ); document.write( "take the square root of both sides of the equation to get:
\n" ); document.write( "\"x+=+3\"
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\n" ); document.write( "that's your answer.
\n" ); document.write( "it was not a simple problem.
\n" ); document.write( "you needed to be able to convert bases: \"log%284%2Cb%29+=+log%282%2Cb%29%2Flog%282%2C4%29\"
\n" ); document.write( "you also needed to remember: \"log%282%2Ca%29+%2B+log%282%2Cb%29+=+log%282%2Ca%2Ab%29\"
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\n" ); document.write( "fyi:
\n" ); document.write( "you could have solved \"log%282%2C4%29\" by converting to \"log%2810%2C4%29%2Flog%2810%2C2%29\" as well as the way i showed you above. then you could have used the log function of the calculator to get your answer.
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