Question 594729
ln(x+2)-ln(x-1)=1
Solving equations like this, where the variable is in the argument (or base) of a logarithm usually starts with transforming the equation into one of the following forms:
log(expression) = other-expression
or
log(expression) = log(other-expression) -- where the bases of the two logs are equal.<br>
With the constant term of 1 (on the right side) the "all log" second form will be challenging to achieve. So we are going to transform your equation into the first form.<br>
The first form has a single logarithm. Your equation has two logs. Somehow we must combine the two logarithms into one. They are not like terms so we cannot simply subtract them. Fortunately there is a property will we can use:
{{{log(a, (p)) - log(a, (q)) = log(a, (p/q))}}}
Using this property on your equation we get:
{{{ln((x+2)/(x-1))=1}}}
We now have the equation in the first form.<br>
Once the equation is in the first form, the next step is to rewrite the equation in exponential form. In general
{{{log(a, (p)) = q}}}
is equivalent to
{{{a^q = p}}}
Using this pattern on your equation we get:
{{{e^1 = (x+2)/(x-1)}}}
which simplifies to:
{{{e = (x+2)/(x-1)}}}<br>
Notice how the variable is now "exposed". We can now use standard algebra to solve this equation for x. First let's eliminate the fraction by multiplying each side by (x-1):
{{{(x-1)*e = (x-1)*((x+2)/(x-1))}}}
which simplifies to:
{{{e*x-e = x+2}}}<br>
Next we will gather the "x-terms" on one side and the "non x-terms" on the other side of the equation. Subtracting e*x and 2 from each side we get:
{{{-e-2 = x - e*x}}}
Factoring out x on the right side:
{{{-e-2 = x(1 - e)}}}
Then we divide bot sides by (1 - e):
{{{(-e-2)/(1 - e) = x}}}
And we are finished. (If you need a decimal approximation, then replace the e's with 2.71828183 (or some rounded off version of that number) and simplify.)