Question

How do you solve `lnx+ln(x-2)=1`?

Answer 1

I found: `x=1+sqrt(1+e)`

Explanation:

We can use the property of logs that allows us to change the sum into a product of arguments as:
`ln[x*(x-2)]=1`
then use the definition of log, applied to our natural log, to write:
`x(x-2)=e^1`
`x^2-2x-e=0`
solve for `x` to get:
`x_(1,2)=(2+-sqrt(4+4e))/2=(2+-2sqrt(1+e))/2`
`x_1=1+sqrt(1+e)`
`x_2=1-sqrt(1+e)` NOT (because negative)

Answer 2

Use the properties:

`lna+lnb=ln(a*b)`

`lnx=y=>x=e^y`

Then solve the quadratic equation. Answer is:

`x_1=2.928`

`x_2=-0.928`

Explanation:

`lnx+ln(x-2)=1`

Since `lna+lnb=ln(a*b)`

`ln(x*(x-2))=1`

To add an `ln` function to the right side we use `x=lne^x`

`ln(x*(x-2))=lne^1`

`lnx` is a `1-1` function:

`x*(x-2)=e`

`x^2-2x-e=0`

Solving the quadratic:

`Δ=(-2)^2-4*1*(-e)=4+4e`

`x_(1,2)=(-(-2)+-sqrt(4+4e))/(2*1)`

From the calculator `sqrt(4+4e)=3.857`

`x_1=2.928`

`x_2=-0.928`