Answer :
when you add ln, you multiply
ln((5x-1)(x-6)=ln6 do now FOIL
ln(5x^2-30x-x+6)=ln6
therefore
5x^2-31x+6=6 and
5x^2-31x=0 furthermore
x(5x-31)=0
one solution is 0, which we will reject
the other solution is 5x=31
x=31/5
ln((5x-1)(x-6)=ln6 do now FOIL
ln(5x^2-30x-x+6)=ln6
therefore
5x^2-31x+6=6 and
5x^2-31x=0 furthermore
x(5x-31)=0
one solution is 0, which we will reject
the other solution is 5x=31
x=31/5
ln(5x - 1) + ln(x - 6) = ln(6)
ln((5x - 1)(x - 6)) = ln(6)
ln(5x(x - 6) - 1(x - 6) = ln(6)
ln(5x(x) - 5x(6) - 1(x) - 1(-6)) = ln(6)
ln(5x² - 30x - x + 6) = lm(6)
ln(5x² - 31x + 6) = ln(6)
5x² - 31x + 6 = 6
5x² - 31x + 6 - 6 = 6 - 6
5x² - 31x + 0 = 0
x = -(-31) +/- √((-31)² - 4(5)(0))
2(5)
x = 31 +/- √(961 - 0)
10
x = 31 +/- √(961)
10
x = 31 +/- 31
10
x = 31 + 31 x = 31 - 31
10 10
x = 62 x = 0
10 10
x = 6²/₅ x = 0
ln((5x - 1)(x - 6)) = ln(6)
ln(5x(x - 6) - 1(x - 6) = ln(6)
ln(5x(x) - 5x(6) - 1(x) - 1(-6)) = ln(6)
ln(5x² - 30x - x + 6) = lm(6)
ln(5x² - 31x + 6) = ln(6)
5x² - 31x + 6 = 6
5x² - 31x + 6 - 6 = 6 - 6
5x² - 31x + 0 = 0
x = -(-31) +/- √((-31)² - 4(5)(0))
2(5)
x = 31 +/- √(961 - 0)
10
x = 31 +/- √(961)
10
x = 31 +/- 31
10
x = 31 + 31 x = 31 - 31
10 10
x = 62 x = 0
10 10
x = 6²/₅ x = 0
