you can factor it, does this help? It's been forever since I did stuff with logs
(1+ln(x))(2+ln(x)) = 0
Not sure how to handle this problem. Any help would be appreciated.
ln^2(X) + ln(X)^3 + 2 = 0
In the first term the exponent is with the natural log. In the second term the exponent is with the X value.
I know it can be written like this:
ln^2(X) + 3ln(X) + 2 = 0
But I don't know how to deal with the first term.
If it helps, the answers are supposed to be x=e^-1 or x=e^-2.
yeah, it definitely does factor. I think the squared log on the first term can be viewed as the first term of a polynomial to the second power...so i'm going with something like:
let ln^2(X) = t^2
let ln(X)^3 = 3t
t=-1 or -2
Now I'm wondering if there is a more direct way to deal with the natural log form...
So for the first factor;:
1+ln(x) = 0 OR ln(x) = -1
e^(ln(x)) = e^(-1)
x = e^(-1)
For the second same thing:
e^(ln(x)) = e^(-2)
e^(lnx)) = x is the golden rule
khalid is a math professor
I took so many math courses (some as my electives... what was I thinking ), I was only 3 away from a BS or BA in mathematics, can't remember which.
I have forgotten most of the math I learned and I never used it in the real world. Getting a Computer Science degree prepares you more for further work in academia than the real world, imo.
Wolfram alpha says to eliminate the quadratic term by substituting x = 1/3 + log(x)
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