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

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Answer 1 · 2015-12-17T17:00:57

$x= 12$

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 $-10$

$-12 + 2 = -10$

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