Question

How do you solve `Log(x+10) - log(x+4) = log x`?

Answer 1

Apply properties of logarithms and solve the resulting quadratic equation to find
`x = 2` or `x = -5`

Explanation:

We will use the following properties:

• `log(x)-log(y)=log(x/y)`

• `e^log(x) = x`

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`log(x+10) - log(x+4) = log(x)`

`=> log((x+10)/(x+4)) = log(x)`

`=> e^log((x+10)/(x+4)) = e^log(x)`

`=> (x+10)/(x+4) = x`

`=> x+10 = x(x+4) = x^2 + 4x`

`=> x^2 + 3x - 10 = 0`

`=> (x+5)(x-2) = 0`

`=> x = 2` or `x = -5`