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

##### 1 Answer

#### Explanation:

Given:

**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

#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

#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

** 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#