I need help with a few math problems. 1) e^e=2 and 2) ln(x)+ln(x-1)=1?

Question

by the way i have to solve for x. so the first one really is e^(e^x)=2 I have what i think are answers. For the first one, i have x=ln(ln2) or -.3665 and for number 2 i have x=e and x=e+1

Answer 1

You're right on the first one:

e^(e^x) = 2

ln(e^(e^x)) = ln2

e^x = ln2

e^x = ln2

ln(e^x) = ln(ln2)

x = ln(ln2) = approx -0.366512921



For the second one:

ln(x) + ln(x-1) = 1

Combine them:

ln( x(x-1) ) = 1

ln(x²+x) = 1

e^ln(x²-x) = e^1 = e

x² - x = e

x² - x - e = 0

quadratic formula:

x = (1/2) (1 ± √(1+4e) )

approx:
x = 2.22287
x = -1.22287

NOTE: this means you may have to exclude the second value (when ± is - ) because unless you're using complex analysis (college level stuff), you should not be taking the ln of negative x.

x = (1/2) (1 + √(1+4e) ) = approx 2.22287

Answer 2

Your first answer is right.

For the second you need to rewrite the equation:
ln(x(x-1)) = 1 = ln(e)

so x^2 - x - e = 0. You can solve the quadratic equation
I believe the roots are 1/2 +/- sqrt(1/4 + e)

Answer 3

1) e^(e^x) = 2 =>
e^x = ln 2 =>
x = ln ln 2

ln(x) + ln(x - 1) = 1

ln(x - 1) = 1 - ln(x)

ln(x - 1) = ln(e) - ln(x)

ln(x - 1) = ln(e/x)

x - 1 = e/x

x^2 - x = e

x^2 - x - e = 0

x1,2 = [1 +- sqrt(1 + 4e)]/2

x1 = 1/2 + sqrt(1 + 4e)/2

x2 = 1/2 - sqrt(1 + 4e)/2 <0, not a real solution

x = 1/2 + sqrt(1 + 4e)/2

Answer 4

e^(e^x) = 2
e^x = ln(2)
x = ln(ln(2))

ln(x)+ln(x-1)=1
ln (x^2-x) +1
x^2-x = e
x = [1 +/- sqrt(1+4e)]/2