Solving for x. ln(x) + ln(x-2) = ln4

Question

Solving for x. ln(x) + ln(x-2) = ln4

no calculator. All by hand

Answer 1

(e^ln(x)+ln(x-2)) = (e^ln(4))

e^ln(x*(x-2)) = e^ln(4)

e^ln(x² - 2x) = e^ln(4)

x² - 2x = 4

x² - 2x - 4 = 0

So then you can use the quadratic equation x = (-b ± sqrt(b² - 4ac)) / 2a), where a = 1, b = -2 and c = -4. you find the values of a, b, and c by looking at your polynomial => ax²+bx+c, 1x² - 2x - 4 = 0 thus a = 1, b = -2, and c = -4.

Then you plug in the values in to the quadratic equation to get your values of x.

x = (-b + sqrt(b² - 4ac)) / 2a)

= (-1*-2 + sqrt((-2)² - 4*1*-4))/(2*1)

= (2+sqrt(4+16))/2 = (2+sqrt(20))/2

= (2+4.4721) / 2

= 6.4721 / 2

= 3.2361

x = (-b - sqrt(b² - 4ac)) / 2a)

= (-1*-2 - sqrt((-2)² - 4*1*-4))/(2*1)

= (2-sqrt(4+16))/2 = (2+sqrt(20))/2

= (2-4.4721) / 2

= -2.4721 / 2

= -1.2361

So x could be 3.2361 or -1.2361, so then you just have to plug in these values for x and see which ones make the equation true. In this case the only value that works is 3.2361.

Answer 2

If you use properties of logarithms like

ln(A) + ln(B) = ln(AB)

and

ln(A) = ln(B) ==> A = B

you will be able to solve this as a quadratic without using any calculus at all (or any calculator either).