document.write( "Question 818511: Consider the function f(x) =log2(2/x^2-3x+2). For what value(s) does f(x)=0 \n" ); document.write( "
Algebra.Com's Answer #492621 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
First of all, I'm assuming the 2 next to the log is an indication of the base of the log. In the future either use some English to make it clear (like \"base 2 log of (2/(x^2-3x+2))\" or teach yourself algebra.com's formula syntax. (Click on the \"Show source\" link above to see what I typed (and what you could type) to make the logarithms look good.)

\n" ); document.write( "Second, I'm assuming that the function is:
\n" ); document.write( "\"f%28x%29+=log%282%2C%282%2F%28x%5E2-3x%2B2%29%29%29\"
\n" ); document.write( "and not:
\n" ); document.write( "\"f%28x%29+=log%282%2C%282%2Fx%5E2-3x%2B2%29%29\"
\n" ); document.write( "which is what you posted. If I am right then please put parentheses around multiple-term numerators, denominators, exponents, function arguments, etc. so that the expression means what you intend it to mean.

\n" ); document.write( "Third, please post your questions in an appropriate category. You posted this under Polynomials. But this problem has a lot more to do with logarithms than polynomials. Posting your question in an appropriate category will increase your chances of a speedy response.

\n" ); document.write( "\"f%28x%29+=log%282%2C%282%2F%28x%5E2-3x%2B2%29%29%29\"
\n" ); document.write( "f(x) = 0
\n" ); document.write( "Replacing f(x) with 0:
\n" ); document.write( "\"0+=log%282%2C%282%2F%28x%5E2-3x%2B2%29%29%29\"
\n" ); document.write( "Next we rewrite this in exponential form:
\n" ); document.write( "\"2%5E0+=+2%2F%28x%5E2-3x%2B2%29\"
\n" ); document.write( "which simplifies to:
\n" ); document.write( "\"1+=+2%2F%28x%5E2-3x%2B2%29\"
\n" ); document.write( "Now that the variable is out of the logarithm we can solve the equation. Multiplying both sides by \"x%5E2-3x%2B2\" (to eliminate the fraction) we get:
\n" ); document.write( "\"x%5E2-3x%2B2+=+2\"
\n" ); document.write( "This is a quadratic equation so we want one side to be zero. Subtracting 2 from each side we get:
\n" ); document.write( "\"x%5E2-3x+=+0\"
\n" ); document.write( "Now we factor:
\n" ); document.write( "\"x%28x-3%29+=+0\"
\n" ); document.write( "Next the Zero Product Property:
\n" ); document.write( "x = 0 or x-3 = 0
\n" ); document.write( "Solving the second equation we get:
\n" ); document.write( "x = 0 or x = 3

\n" ); document.write( "Last we check. This is not optional! A check must be made to ensure that all bases and arguments of all logs are valid. (Valid bases are positive but not 1 and valid arguments are positive.) If a \"solution\" makes any base or argument invalid it must be rejected.

\n" ); document.write( "Use the original equation to check:
\n" ); document.write( "\"0+=log%282%2C%282%2F%28x%5E2-3x%2B2%29%29%29\"
\n" ); document.write( "Checking x = 0:
\n" ); document.write( "\"0+=log%282%2C%282%2F%28%280%29%5E2-3%280%29%2B2%29%29%29\"
\n" ); document.write( "Simplifying:
\n" ); document.write( "\"0+=log%282%2C%282%2F%280-3%280%29%2B2%29%29%29\"
\n" ); document.write( "\"0+=log%282%2C%282%2F%280-0%2B2%29%29%29\"
\n" ); document.write( "\"0+=log%282%2C%282%2F2%29%29%29\"
\n" ); document.write( "We can now see that the base, 3, and the argument, 2/2 (or 1), are both valid. So this solution passes the check.

\n" ); document.write( "Checking x = 3:
\n" ); document.write( "\"0+=log%282%2C%282%2F%28%283%29%5E2-3%283%29%2B2%29%29%29\"
\n" ); document.write( "Simplifying:
\n" ); document.write( "\"0+=log%282%2C%282%2F%289-3%283%29%2B2%29%29%29\"
\n" ); document.write( "\"0+=log%282%2C%282%2F%289-9%2B2%29%29%29\"
\n" ); document.write( "\"0+=log%282%2C%282%2F2%29%29%29\"
\n" ); document.write( "Again this solution checks out.

\n" ); document.write( "So there are two solutions to your equation: x = 0 or x = 3. \n" ); document.write( "
\n" );