Question

How do you solve `ln x + ln (x+3) = 1`?

Answer 1

Combine the left into one term using the rule `ln(ab)=lna+lnb`

`lnx+ln(x+3)=1`
`ln[(x)(x+3)]=1`
`ln(x^2+3x)=1`

Rewrite using the definition of log:
`e^1=x^2+3x`
`e=x^2+3x`

Now solve as a quadratic; first set one side equal to zero:
`0=x^2+3x-e`

Use quadratic formula:
`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(-(3)+-sqrt(3^2-4(1)(e)))/(2(1))`
`x=(-3+-sqrt(9-4e))/2`

`xapprox(-3+-sqrt(-1.87))/2`

Because we take the square root of a negative number, there are no solutions to the equation.