Let $a, b\in\mathbb{R}^*_+$ and $p, q\in\mathbb{R}, p, q>1$. I would like to find the right exponent $x$ (clearly $x$ depending on $p$ and $q$), such that $$ a^p\le (a+b)^x \quad\mbox{ and }\quad b^q\le (a+b)^x.$$
Could someone please suggest some possible options? Every hints or idea is well accepted.
Thank you in advance!
$\bf EDIT:$ for the case $a+b>1$, I think that the answer of Gareth Ma works. What about the case $0<a+b<1$? Thanks.
Taking logarithms, the equations give $p\log a\leq x\log(a + b)$ and $q\log b\leq x\log(a + b)$. Therefore, we need (assuming $a + b > 1$)
$$\begin{cases} x&\geq \frac{1}{\log(a + b)} \max\{p\log a, q\log b\} &\text{if}\ a + b > 1 \\ x &\leq \frac{1}{\log(a + b)} \min\{p\log a, q\log b\}&\text{if}\ a + b < 1\end{cases}$$