Question 365204
I assume the expression is:
{{{ln(x/(x-2))+ln((x+2)/x)-ln(x^2-4)}}}
These are not like terms because the arguments are different. But there are properties of logarithms which allow us to combine these logarithms into one:<ul><li>{{{log(a, (p)) + log(a, (q)) = log(a, (p*q))}}}</li><li>{{{log(a, (p)) - log(a, (q)) = log(a, (p/q))}}}</li></ul>
With these properties all you need is the same base and coefficients of 1 and your logarithms meet both requirements. Using the first property on the first two logarithms (because of the "+" between them) we get:
{{{ln(x/(x-2))*((x+2)/x)-ln(x^2-4)}}}
The x's cancel:
{{{ln(cross(x)/(x-2))*((x+2)/cross(x))-ln(x^2-4)}}}
leaving:
{{{ln((x+2)/(x-2))-ln(x^2-4)}}}
Now we can use the second property (because of the "-" between them):
{{{ln(((x+2)/(x-2))/(x^2-4))}}}
Now we try to simplify the compllex fraction. We start by factoring the denominator:
{{{ln(((x+2)/(x-2))/((x+2)(x-2)))}}}
Now we will multiply the numerator and denominator by (x-2):
{{{ln((((x+2)/(x-2))/((x+2)(x-2)))((x-2)/(x-2)))}}}
The (x-2)'s in the numerator cancel:
{{{ln((((x+2)/cross((x-2)))/((x+2)(x-2)))(cross((x-2))/(x-2)))}}}
leaving:
{{{ln((((x+2)/1)/((x+2)(x-2)))(1/(x-2)))}}}
which simplifies to:
{{{ln((x+2)/((x+2)(x-2)(x-2)))}}}
Now the (x+2)'s cancel:
{{{ln(cross((x+2))/(cross((x+2))(x-2)(x-2)))}}}
leaving:
{{{ln(1/((x-2)(x-2)))}}}
Since it says not to use powers, we will not change the denominator to {{{(x-2)^2}}}. This may be an acceptable answer. However we can simplify a little further. We can rewrite the argument as follows:
{{{ln(((x-2)(x-2))^(-1))}}}
and then use the thrid property of logarithms, {{{log(a, (p^q)) = q*log(a, (p))}}, to move the exponent out in front:
{{{(-1)*ln((x-2)(x-2)))}}}
or
{{{-ln((x-2)(x-2)))}}}