Question 818511
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.)<br>
Second, I'm assuming that the function is:
{{{f(x) =log(2,(2/(x^2-3x+2)))}}}
and not:
{{{f(x) =log(2,(2/x^2-3x+2))}}}
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.<br>
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.<br>
{{{f(x) =log(2,(2/(x^2-3x+2)))}}}
f(x) = 0
Replacing f(x) with 0:
{{{0 =log(2,(2/(x^2-3x+2)))}}}
Next we rewrite this in exponential form:
{{{2^0 = 2/(x^2-3x+2)}}}
which simplifies to:
{{{1 = 2/(x^2-3x+2)}}}
Now that the variable is out of the logarithm we can solve the equation. Multiplying both sides by {{{x^2-3x+2}}} (to eliminate the fraction) we get:
{{{x^2-3x+2 = 2}}}
This is a quadratic equation so we want one side to be zero. Subtracting 2 from each side we get:
{{{x^2-3x = 0}}}
Now we factor:
{{{x(x-3) = 0}}}
Next the Zero Product Property:
x = 0 or x-3 = 0
Solving the second equation we get:
x = 0 or x = 3<br>
Last we check. This is <i>not</i> 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.<br>
Use the original equation to check:
{{{0 =log(2,(2/(x^2-3x+2)))}}}
Checking x = 0:
{{{0 =log(2,(2/((0)^2-3(0)+2)))}}}
Simplifying:
{{{0 =log(2,(2/(0-3(0)+2)))}}}
{{{0 =log(2,(2/(0-0+2)))}}}
{{{0 =log(2,(2/2)))}}}
We can now see that the base, 3, and the argument, 2/2 (or 1), are both valid. So this solution passes the check.<br>
Checking x = 3:
{{{0 =log(2,(2/((3)^2-3(3)+2)))}}}
Simplifying:
{{{0 =log(2,(2/(9-3(3)+2)))}}}
{{{0 =log(2,(2/(9-9+2)))}}}
{{{0 =log(2,(2/2)))}}}
Again this solution checks out.<br>
So there are two solutions to your equation: x = 0 or x = 3.