Question

Solve for `x` when `log(x^2-x-6)+x=log(x+2)+4`?

Answer 1

`x=4`

Explanation:

`log(x^2-x-6)=log((x-3)(x+2))`
`=log(x-3)+log(x+2)`

Therefore,
`log(x-3)+log(x+2)+x=log(x+2)+4`
`log(x-3)=4-x`
`x-3=10^(4-x)`

Solutions to problems such as these cannot be calculated via regular methods, however, just looking at this equation, the answer can be seen to be `4`, as `4-3` is `1`, and `10^0` is also `1`.

Answer 2

`x=4`

Explanation:

`log(x^2-x-6)+x=log(x+2)+4`

Move all the `log` functions to one side and the constants the other,

`log(x^2-x-6)-log(x+2)=4-x`

Make use of the laws of logarithms,

`log((x^2-x-6)/(x+2))=4-x`
`color(white)(xxxxx)log(x-3)=4-x`

Since `log(x-3)` and `4-x` are equal, plot them against `y`,

`color(purple)(y=log(x-3)`
`y=4-x`

Hence, `x=4`.