Question 681864
{{{log(2,(x-6)) +log(2,(x-4)) - log( 2, (x) ) =2}}}
By "condense it and put x as the denominator" I hope you mean...
Using {{{log(a, (p)) + log(a, (q)) = log(a, (p*q))}}} to combine the first two logs:
{{{log(2,((x-6)(x-4))) - log( 2, (x) ) =2}}}
which simplifies to:
{{{log(2,(x^2-10x+24)) - log( 2, (x) ) =2}}}
Using {{{log(a, (p)) - log(a, (q)) = log(a, (p/q))}}} to combine the remaining logs:
{{{log(2,((x^2-10x+24)/x)) =2}}}<br>
The next step is to rewrite the equation in exponential form. In general {{{log(a, (p)) = k}}} is equivalent to {{{a^k = p}}}. Using this pattern on our equation we get:
{{{2^2 = (x^2-10x+24)/x}}}
which simplifies to:
{{{4 = (x^2-10x+24)/x}}}<br>
Now that the logs are gone we can use "regular" algebra to solve. First let's multiply by x to eliminate the fraction:
{{{4x = x^2-10x+24}}}
Since this is a quadratic equation we want one side to be you. Subtracting 4x we get:
{{{0 = x^2-14x+24}}}
Next we factor or use the Quadratic Equation. This factors fairly easily:
{{{0 = (x-12)(x-2)}}}
From the Zero Product Property:
{{{x - 12 = 0}}} or {{{x - 2 = 0}}}
Solving these we get:
{{{x = 12}}} or {{{x = 2}}}<br>
Last of all, we check our answers. This is <b>not optional</b> when solving these kinds of equations. You must ensure that any "solution" will make all arguments to all logarithms positive. If a "solution" makes any argument zero or negative then we must reject that solution. Use the original equation to check:
{{{log(2,(x-6)) +log(2,(x-4)) - log( 2, (x) ) =2}}}
Checking x = 12:
{{{log(2,((12)-6)) +log(2,((12)-4)) - log( 2, ((12)) ) =2}}}
We can already see that all three arguments will be positive when x = 12. So this solution passes the required part of the check.<br>
Checking x = 2:
{{{log(2,((2)-6)) +log(2,((2)-4)) - log( 2, ((2)) ) =2}}}
We can see that the first two arguments will be negative when x = 2. So this solution fails the check and we reject it.<br>
So there is just one solution to the equation: x = 12