What is the limit of $\log_k(k^a + k^b)$ for $k \to +\infty$? I'm not very good with analysis (I never studied it) but because of my "work" on other topics of mathematics I came to this problem.
$$\lim_{k \to +\infty }\log_k(k^a + k^b)=\max(a,b)$$
I'm sure that this is really "reasonable" because I tryed to graph it for really huge values of $k$ ... but "reasonable" is not enough in mathematics: I'd like to prve this.
 So I'm courious to know how one should go to prove it in a formal way ... if it is true (I hope).
PS: I know nothing about Limits and their rules
 A: Use the functional equation:
$\log(k^a+k^b)=a+\log(1+k^{b-a})$
When $a> b$ the log term goes to 0, if $b\ge a$ it goes to $b-a$ and $b-a+a=b$.
A: Assume that $a\geqslant b$ then $k^a\leqslant k^a+k^b\leqslant 2k^a$ hence $a\leqslant\log _k(k^a+k^b)\leqslant a+\log_k2$ and $\log_k2=\frac{\log2}{\log k}$. Can you finish?
A: If $a\gt b$ then $$\log_k(k^a+k^b)=\log_k k^a+\log_k (1+k^{b-a})$$
(Note that if $a=b$ you get $a+\log_k(2)$)
Then the first term is constant and the limit of the second is easy to establish.
A: I would try the following. Notice that: $$\lim_{k \to + \infty} \log_k(k^a + k^b) = \lim_{k \to + \infty} \frac{\ln(k^a + k^b)}{\ln k}$$
If $0 < a,b < 1$, the limit is $0$. Suppose $a,b > 1$ and use L'Hospital's rule. We get: $$\lim_{k \to +\infty} \frac{\frac{k^a \ln k + k^b \ln k}{k^a + k^b}}{\frac{1}{k}} = \lim_{k \to +\infty} \ln k \frac{k^{a +1} + k^{b + 1}}{k^a + k^b} = \lim_{k \to + \infty} \ln k \frac{k^{a + 1 -b} + k}{k^{a - b} + 1} \\ = \lim_{k \to +\infty} k \ln k \frac{k^{a - b} + 1}{k^{a - b} + 1} = \lim_{k \to +\infty} k \ln k = + \infty$$
A: Without loss if generality assume $a\ge b$ then:
$$\log_{k}(k^{a}+k^{b})=a+\log_{k}(1+k^{b-a})=a+\frac{\ln(1+k^{b-a})}{\ln(k)}$$
where I changed the base of the logarithm to get the second equality. Notice that if $a>b$ then $1+k^{b-a}\to1$ and $\ln(k)\to\infty$ both as $k$ tends to $\infty$. If $a=b$ then $1+k^{b-a}=2$ but $\ln(k)\to\infty$ as $k\to\infty$. Thus, the above tends to $a=\max\{a,b\}$.
