Question

Solve ln(x^2+x)-ln(x^2-x)=-1​

Answer 1

Explanation:

Using the following property of logarithms,

`\ln{\left((a)/(b)\right)} = ln(a) - ln(b)`

we can write the given equation as

`ln((x^2 + 1)) - ln((x^2 - 1)) = \ln{\left((x^2 + 1)/(x^2 - 1)\right)}`

`\:\:\:\:\:\:\:\:\:= \ln{(x(x + 1))/(x(x - 1))} = \ln{\left((x + 1)/(x - 1)\right)}`

But recall that
`ln(e) = 1` so we can write our original equation as

`\ln{\left((x + 1)/(x - 1)\right)} = ln(e)`

or simply

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

Multiplying both sides by
`(x - 1),` we get

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

Solving for x, we get

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