Question

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

Answer 1

`color(blue)(x=(-1+sqrt(4e+9))/2~~1.7290)`

Explanation:

Using the logarithmic property:

`lna+lnb=lnab`

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

Using the antilogarithm:

`e^(ln(x-1)(x+2))=e^1`

`(x-1)(x+2)=e`

Expanding `LHS`

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

Using quadratic formula:

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

`x=(-1+sqrt(4e+9))/2~~1.7290`

`x=(-1-sqrt(4e+9))/2~~-2.7290`

Checking solutions with original equation:

`ln(1.7290-1)+ln(1.7290+2)=1.000058554`

`ln(-2.7290-1)+ln(-2.7290+2) \ \ color(red)(X)`

This is undefined for real numbers.

`lnx` is only defined for real numbers if `x>0`

Answer 2

`x=\frac{-1+\sqrt{9+4e}}{2}=1.72896`

Explanation:

`\ln(x-1)+\ln(x+2)=1\ quad (\forall \ \ x>1)`

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

`\ln(x^2+x-2)=\ln e`

Comparing the numbers on same base on both the sides,

`x^2+x-2=e`

`x^2+x-(2+e)=0`

Solving above quadratic equation as follows

`x=\frac{-1\pm\sqrt{1^2-4(1)(-(2+e))}}{2(1)}`

`x=\frac{-1\pm\sqrt{9+4e}}{2}`

But, `x>1` hence

`x=\frac{-1+\sqrt{9+4e}}{2}`

`=1.72896`