# Why can't use natural log for this equation?

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

#### romsek

MHF Helper

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.

#### Plato

MHF Helper
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
Well the logarithm is not an additive function.
$\Large\log(e^{2n}-2e^{n}+6)\ne\log(e^{2n})-\log(e^{n})-\log(6)$

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.

#### romsek

MHF Helper
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.
$\ln(a+b) \neq \ln(a) + \ln(b)$

#### HallsofIvy

MHF Helper
Instead, let $y= e^x$ so the equation becomes $y^2- 2y- 6= 0$. Solve that by completing the square or the quadratic formula then take the logarithm.