Question

`log_2(x-3) = 2- log_2(x-6)` What is x?

Answer 1

`x = 7`

Explanation:

`log_2(x-3) = 2 - log_2(x-6)`
`log_2(x-3) + log_2(x-6) = 2`
`log_2((x-3)(x-6)) = 2`
`log_2(x^2-6x-3x+18) = 2`
`log_2(x^2-9x+18) = 2`
`x^2 - 9x + 18 = 4`
`x^2 -9x +14 = 0`

Now that the equation is in the form `ax^2 + bx + c = 0`, the quadratic formula (`x = (-b+-sqrt(b^2-4ac))/(2a)`) can be applied:
`x = (-(-9)+-sqrt((-9)^2-4(1)(14)))/(2(1))`
`x=(9+-sqrt(81-56))/2`
`x=(9+-sqrt(25))/2`
`x = (9+-5)/2`

Adding `5` for the first root of `x`:
`x = (9+5)/2 = 14/2 = 7`

Subtracting `5` for the second root of `x`:
`x = (9-5)/2 = 4/2 =2`

However, a `log` can not be applied to a negative number, so there are two more conditions for the root(s) of x:
`(x-3)>0` and `(x-6)>0`

Testing `7` to see if it is a root of `x`;
`(7-3)>0`, `4>0` The first condition is met.
`(7-6)>0`, `1>0` the second condition is met.
Thus, `7` is a root of `x`.

Testing `2` to see if it is a root of `x`:
`(2-3)>0` Since `-1` is not more than `0`, `2` is not a root of `x`.

Answer 2

`x=7`

Explanation:

`"using the "color(blue)"laws of logarithms"`

`•color(white)(x)logx+logy=log(xy)`

`•color(white)(x)log_b x=nhArrx=b^n`

`"add "log_2(x-6)" to both sides"`

`rArrlog_2(x-3)+log_2(x-6)=2`

`rArrlog_2(x-3)(x-6)=2`

`rArr(x-3)(x-6)=2^2=4`

`rArrx^2-9x+14=0larrcolor(blue)"in standard form"`

`"the factors of + 14 which sum to - 9 are - 2 and - 7"`

`rArr(x-2)(x-7)=0`

`"equate each factor to zero and solve for x"`

`x-2=0rArrx=2`

`x-7=0rArrx=7`

`(x-3)>0" and "(x-6)>0`

`rArrx=2" is invalid"`

`rArrx=7" is the solution"`