Question

How do you simplify `log (x+6)=1-log(x-5)`?

Answer 1

I tried this:

Explanation:

We can use, as an example, the base of the logs as being `10`. We get:

`log_(10)(x+6)=1-log_(10)(x-5)`

rearrange:

`log_(10)(x+6)+log_(10)(x-5)=1`

use a property of logs to write:

`log_(10)[(x+6)(x-5)]=1`

use the definition of log to write:

`(x+6)(x-5)=10^1`

that becomes:

`x^2-5x+6x-30-10=0`
`x^2+x-40=0`

Use the Quadratic Formula:

`x_(1,2)=(-1+-sqrt(1+160))/2=(-1+-sqrt(161))/2`
and:
`x_1=(-1+sqrt(161))/2=5.8442`
`x_2=(-1-sqrt(161))/2=-6.8442`

the negative solution cannot be accepted.

Answer 2

`x = (-1 + sqrt(161))/2 ~~ 5.844`

Explanation:

`log(x+6) = 1 - log(x-5)`

Move `log(x-5)` to Left Side
`log(x+6) + log(x-5) = 1`

Condense the Left Side into 1 Log expression
`log((x+6)(x-5)) = 1`

Convert from Log Equation to Exponential Equation
`(x+6)(x-5) = 10^1`

Multiply Left Side
`x^2 - x - 30 = 10`

Move `10` to Left Side
`x^2 - x - 40 = 0`

Since we can't factor, use Quadratic Formula:
Quadratic Formula: `x = (-b +- sqrt(b^2 - 4ac))/(2a)`
`x = (-1 + sqrt(161))/2 ~~ 5.844`
And
`x = (-1 - sqrt(161))/2 ~~ -6.844`

Since the "inside" of log has to be positive,
`x = (-1 + sqrt(161))/2 ~~ 5.844` is the answer.