Is it ok to switch to logarithms when testing divergence of sequences involving a sum of exponentials? E.g $x_n = 5^n - 4^n \to \ln5^n - \ln4^n$

Question

Consider the following sequence:

$$ x_n = 5^n - 4^n $$

I'm wondering whether it is ok to switch from comparing exponents to comparing logarithms. Here is what I mean:

Instead of $5^n - 4^n$ compare:

$$ \ln5^n - \ln4^n = n(\ln5 - \ln4) = n\ln{5\over4} $$

Which is clearly divergent. Is that true that if the expression involving logarithms diverges then the expression with exponentials also diverges?

Update

As pointed in the comments it's not very clear what kinds of sequences are taken into consideration. Those are in the form of:

$$ x_n = a_1^n + a_2^n + \dots + a_k^n $$

Where $a_k$ is some constant term and the number of terms is limited by $k$.

Answer 1

$1, 1+1, 1+1+1, ...$ diverges but $\ln 1, \ln 1 + \ln 1, \ln 1 + \ln 1 + \ln 1...$ is converges.

There's really no comparitive relationship between $a + b$ and $\ln a + \ln b$ especially as $a + b = (a+ k) +(b-k)$ and $\ln a + \ln b$ need not have any predictable comparison to $\ln (a+k) + \ln (b-k) = M*\ln a + \frac 1M \ln b$ for any arbitrary constant $M$. As we can manipulate $M$ to get just about any value there's really nothing we can conclude.

Even comparing $a + b$ to $\ln(a + b)$ should be taken with a grain of salt as we have to prove not only that both increasing and decreasing is maintained in conversion but that boundedness is as well.

So as $\ln$ is not distributive attempting any comparison of $a + b$ to $\ln a + \ln b$ should set off many red lights and should not set off any green ones.

Answer 2

This is not ok, you can make this here $$\log(ab)=\log(a)+\log(b)$$ $$\log\left(\frac{a}{b}\right)=\log(a)-\log(b)$$ for $$a,b>0$$