For students who studied the logarithmic function, it is easy to solve the equation $$0.9^n \leq 0.1$$ in $\mathbb{N}$, which has as solutions $n\geq \frac{\ln0.1}{\ln 0.9} \approx 21.85 $. That is all natural numbers starting from $22$.
Now how can we solve the following equation $$0.9^n+0.8^n \leq 0.1$$
On
Asking a numerical solver to find $n$ such that $0.8^n + 0.9^n = 0.1$, yields $n = 22.5020{\dots}$. Checking that the function decreases as $n$ increases, the solution set is $n \geq 23$.
OP has stated they do not want to use this method. But at least we know what answer we should be getting.
Just start trying $n$... Clearly, when $n = 0$, the sum is $2 > 1$ and the sum decreases as $n$ increases. So, start with $n = 1$ and see when you have success. \begin{align*} &n & &0.8^n + 0.9^n \\ &1 & & 1.7 \\ &2 & & 1.45 \\ &3 & & 1.241 \\ &4 & & 1.0657 \\ &5 & & 0.91817 \\ &\vdots & &\vdots \\ &21 & & 0.1186\dots \\ &22 & & 0.10585\dots \\ &23 & & 0.0945 \dots \end{align*} So the solution set in $\Bbb{N}$ is $n \geq 23$.
There are a few ways to compute less of this table.
Notice that $2 \cdot 0.8^n < 0.8^n + 0.9^n < 2 \cdot 0.9^n$. Solving $2 \cdot 0.8^n = 0.1$, we get $n = 13.425{\dots}$. Solving $2 \cdot 0.9^n = 0.1$, we get $n = 28.433{\dots}$. So the solution to the original equation is one of $n \geq 14$, $n \geq 15$, $\dots$, $n \geq 29$. We could check these in order (as in the table above), to find a solution.
However, we can binary search this region, which is much quicker. \begin{align*} &n & &0.8^n + 0.9^n \\ &14 & & 0.272\dots \\ &29 & & 0.0486\dots \\ \left\lfloor \frac{14+29}{2} \right\rfloor &= 21 & & 0.1186\dots \\ \left\lfloor \frac{21+29}{2} \right\rfloor &= 25 & & 0.07556\dots \\ \left\lfloor \frac{21+25}{2} \right\rfloor &= 23 & & 0.0945\dots \\ \left\lfloor \frac{21+23}{2} \right\rfloor &= 22 & & 0.1058\dots \end{align*} and we find that $n \geq 23$ is the solution.
You have shown that $0.9^n \leq 0.1$ when $n \geq 22$. Since $0.8^n > 0$, $0.8^n + 0.9^n > 0.9^n$, so we can shortcut the list in answer one by starting at $n = 22$, since the sum can't be smaller than $0.1$ if the larger term is not. This leads us to compute only the last two rows in the table.
On
Quanto's solution is a good practical solution because, if we generalise the problem so 0.1 can be smaller, say replace 0.1 with 0.001, then using the same method we see that 0.9^n <= 0.001 implies n >= 66, and the fact that 0.8^66 is much smaller than 0.9^66, we find that 0.8^n has basically no effect and in fact the answer is n >= 66.
However, what if the problem was, solve: 0.9^n + 0.899^n <= 0.001 ? where the 0.899^n probably does have some effect on the final answer?
Then we get roughly: 2 x 0.9^n <= 0.001 which gives n >= 73, so our actual solution n to the above will be less than 73. But we already know that n >= 66, therefore our solution will be n >= p where 66 <= p <= 73, and we have fairly minimal guesswork to find the solution (which, if you're curious, is n >= 72).
I'm not sure if/when we get a large enough gap between our lower bound and our upper bound in some similar question so that guesswork becomes too much work.
On
Using whole numbers, you are looking for the zero of $$f(x)=\left(\frac{9}{10}\right)^x+\left(\frac{4}{5}\right)^x-\frac{1}{10}$$ and, as said in comments and answers, a numerical method should be required.
If you plot the function, it is not very nice while $$g(x)=\log\left(\left(\frac{9}{10}\right)^x+\left(\frac{4}{5}\right)^x \right)+\log(10)$$ looks like a straight line.
Being very lazy, using Taylor series at $x=0$ would give $$g(x)=\log (20)+ \log \left(\frac{3 \sqrt{2}}{5}\right)x+O\left(x^2\right)$$ which would provide as an estimate $x=-\frac{\log (20)}{\log \left(\frac{3 \sqrt{2}}{5}\right)}\approx 18.24$.
Being less lazy, we can notice that $f(x)$ is bracketed by $2\left(\frac{9}{10}\right)^x-\frac{1}{10}$ and $2\left(\frac{4}{5}\right)^x-\frac{1}{10}$ which gives as bounds $13.43$ and $28.42$.
Using Newton method with $x_0$ at the mid point of the interval would give the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 20.9291 \\ 1 & 22.4918 \\ 2 & 22.5021 \end{array} \right)$$
From the solution to the previous problem $0.9^n\le 0.1$, we know $n\ge22$.
Now, try,
$$0.9^{22 }+ 0.8^{22 }-0.1=0.006$$
$$0.9^{23}+0.8^{23}-0.1= -0.005$$
The values switch signs from 22 to 23. So, from the intermediate-value-theorem, the solution is
$$n\ge23$$