Question

log(x-2) + log(x-3) = 1 + log(2x-7)

Answer 1

log(x-2) + log(x-3) = 1 + log(2x-7)
log[(x-2).(x-3)] = log 10 + log(2x-7)
log(x²-5x+6) = log[10.(2x-7)]
x²-5x+6 = 20x-70
x²-5x-20x+6+70=0
x²-25x+76=0
/\ = 625 - 304
/\ = 321
x = (-(-25)+/- \/321)/2
x = (25+/- \/321)/2
x' = (25+\/321)/2
x" = (25-\/321)/2