Question

How do you solve `log_(5) (x-1) + log_(5) (x-2) - log_(5) (x+6) = 0`?

Answer 1

Use logarithm rules for adding and subtracting logarithms and simplify first before solving. The step by step working is given below.

Explanation:

`log_5(x-1) + log_5(x-2)-log_5(x+6) =0`

Note: `log(1) = 0` for any base.

`log_5(x-1)+log_5(x-2)-log_5(x+6)=log_5(1)`
`log_5(((x-1)(x-2))/(x+6)) = log_5(1)`

Note: `log_b(P) + log_b (Q) = log_b(PQ)` and `log_b(P) - log_b(Q) = log_b(P/Q)`

`((x-1)(x-2))/(x+6) = 1`

Now to solve the equation.

Start by cross multiplying.

`(x-1)(x-2)=(x+6)` As you can see this removes the denominator and makes it easier for us to solve.

Now simplify
`(x-1)(x-2) = x(x-2)-1(x-2)`
`(x-1)(x-2)=x^2-2x-x+2`
`(x-1)(x-2)=x^2-3x+2`

Our problem now becomes

`x^2-3x+2 = x+6`
Subtracting `x+6` from both the sides we get.
`x^2-3x+2-x-6=0`
`x^2-4x-4=0`

Solving for `x` using the quadratic formula.

Quadratic formula
`x=(-b+-sqrt(b^2-4ac))/(2a)`
Here `a=1`, `b=-4` and `c=-4`
`x=(-(-4)+-sqrt((-4)^2-4(1)(-4)))/(2(1))`
`x=(4+-sqrt(16+16))/2`
`x=(4+-sqrt(16*2))/2`
`x=(4+-4sqrt(2))/2`
`x=(2+-2sqt(2))`

`x=2-2sqrt(2)` will give a negative number and won't satisfy the given equation. Therefore the solution is `x=2+2sqrt(2)`