Question

How do you solve `log ( x -3 ) + log ( x-5 ) = log ( 2x - 9)`?

Answer 1

`x= 12`

Explanation:

Given:

`log(x-3) + log(x-5) = log(2x-9)`

Step 1: Rewrite the expression using sum to product rule

`log[(x-3)(x-5)] = log(2x-9)`

`log(x^2 -3x-5x+15) = log(2x-9)`

Step 2 : Rewrite in exponential form with base to ("drop" log since we have sam log both side of equation)

`10^(log(x^2-8x+15)) = 10^(log(2x-9))`

`x^2 - 8x+15 = 2x-9`

Step 3: Manipulate equation to write it in quadratic form `ax^2 + bx+c= 0`

`x^2 - 8x + 15= 2x- 9`

`x^2 -10x + 24= 0`

Step 4: This can be solve by factoring

`(x-12)(x+2) = 0` **

`color(red)(x-12 = 0 => x= 12`

`x+2 = 0 => x = -2`

Step 5: Check solution- can't have negative number as argument for the logarithm

Check `x= -2`

`log(-2-3) + log(-2-5) = log (2*-2-9)`

`log(-5) + log(-7) = log(-13)`

Can't have negative as argument for logarithm, therefore `x= -2` is extraneous solution

** 2 number multiply equal to 24

like `-12*2 or -6* -4 or -8 *3 or -2*12 " " etc.`
**Add equal to `-10`

`-12 + 2 = -10`