solve ln(x^2-2x)= 1?

Answer 1

ln (x² - 2x) = 1

We know that ln b = log_e b where e is the "euler's number" (approx. 2.71828182845904523536...). If we convert the logarithm statement, we arrive at
e^1 = x² - 2x

Or...
x² - 2x - e = 0

By the quadratic formula,
x = [2 ± √(4 + 4e)]/2

We can take the sqrt 4 = 2
x = [2 ± 2√(e + 1)]/2

We can divide 2
x = 1 ± √(e + 1)

Thus, we have 2 solutions:
x = 1 + √(e + 1) or
x = 1 - √(e + 1)

Check...
ln(x² - 2x) = 1
ln [(1 + √(e + 1))² - 2(1 + √(e + 1))] = 1
ln (1 + 2√(e + 1) + e + 1 - 2 - 2√(e + 1)) = 1
ln e = 1
Thus,
1 = 1

For the other root...
ln [(1 - √(e + 1))² - 2(1 - √(e + 1))] = 1
ln (1 - 2√(e + 1) + e + 1 - 2 + 2√(e + 1)) = 1
ln e = 1
Thus,
1 = 1

^_^

Answer 2

to "eliminitate" ln you have to do e^ln :

you do it each side of the equation:

e^ [ ln(x^2-2x) ] = e^1
x^2 - 2x = e
x^2 - 2x - e = 0

using the quadratic formula, you find :

x = 1 - (e + 1)^½
x = 1 + (e +1)^½

Good luck !

Answer 3

ln(x^2-2x)=1

x^2-2x=e

x(x-2)-e=0

then just plug it into the quadratic equation

Answer 4

x^2-2x=e

so
x= (2 +- sqrt(4-4(-e)) /2
= (2 +- sqrt(4+4e))/2
=(2+- 2sqrt(e))/2

x= 1+e^{1/2}
or
x= 1-e^{1/2}

Answer 5

Let's try this: