SOLUTION: #1) log4(x) + log4(x-3) = 1 This is what I've tried: x(x-3) = 1 x^2 - 3x -1 = 0 ----> I can't factorise so do I need to complete the square to find x or is there an easier way o

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: #1) log4(x) + log4(x-3) = 1 This is what I've tried: x(x-3) = 1 x^2 - 3x -1 = 0 ----> I can't factorise so do I need to complete the square to find x or is there an easier way o      Log On


   

Question 1095569: #1) log4(x) + log4(x-3) = 1
This is what I've tried:
x(x-3) = 1
x^2 - 3x -1 = 0 ----> I can't factorise so do I need to complete the square to find x or is there an easier way of doing it?
#2) log10(2x^2+3x+3) - log10(1-2x) = 1
I'm having the same factorising issue here.
Thankyou so much
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Is the base of log, 4 ?

log%284%2Cx%29%2Blog%284%2C%28x-3%29%29=1

log%284%2C%28x%28x-3%29%29%29=1

4%5E1=x%28x-3%29

x%5E2-3x-4=0, and you can take the rest from here.
You want the logarithms of non-negative numbers.