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#