Question

How do you solve `lnx+ln(x-2)=3`?

Answer 1

We use the fact that `lna+lnb=lnab`

Explanation:

Hence we have that

#lnx+ln(x-2)=3=>ln(x(x-2))=3=> x(x-2)=e^3=> x^2-2x-e^3=0#

The last is a trinomial with respect to x that has the following solutions

`x_1=1-sqrt(1+e^3),x_2=1+sqrt(1+e^3)`

Now we must not forget that `x>2` so the only acceptable solution here is `1+sqrt(1+e^3)`

Answer 2

Try this:

Explanation:

We can use the rule of the logs that says:
`logx+logy=log(xy)` to get:

`ln[x(x-2)]=3`
apply the definition of log (natural, with base `e`) to get:
`x(x-2)=e^3`
rearranging:
`x^2-2x-e^3=0`

Use the Quadratic Formula to solve for `x` as:
`x_(1,2)=(2+-sqrt(4+4e^3))/2`
`x_(1,2)=(2+-sqrt(4(1+e^3)))/2`
`x_(1,2)=(2+-2sqrt(1+e^3))/2`
`x_(1,2)=1+-sqrt(1+e^3)`
Now check whether the two solutions are allowed or not by substituting them into the original equation (you'll find that only one works).