So I have solved the equation below the proper way but cannot understand why applying the natural logarithm this way is wrong when it does work for other equations, what rule I am I not following here ?
https://imgur.com/a/cbOVa
So I have solved the equation below the proper way but cannot understand why applying the natural logarithm this way is wrong when it does work for other equations, what rule I am I not following here ?
https://imgur.com/a/cbOVa
egad...
logs don't work at all like that
what you could do is
$e^{2n} - 2e^n = 6$
$\ln\left(e^{2n} - 2e^n \right) = \ln(6)$
but this is as far as you can go using this strategy.
The actual way to solve this problem is with the substitution
$u = e^n$
Then we can rewrite the original equation as
$u^2 - 2u - 6 = 0$
Solve this for $u$
and then $x = \ln(u),~u>0$
If there are any solutions for $u$ such that $u\leq 0$ they can be ignored.
Wow I went through like 50 questions without a problem and I'm surprised how not knowing this didn't bite me. So if I'm understanding correctly, I can't take the log of every single term separately in an equation.