Solve $0.9^n+0.8^n \leq 0.1$ 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
$$0.9^n+0.8^n \leq 0.1$$
 A: Ask a computer
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.
Basic Answer
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.
Upper and lower bounding with binary search
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.
Using solution to partial equation
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.
A: 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
$$n\ge23$$
A: 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.
A: 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 
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 20.9291 \\
 1 & 22.4918 \\
 2 & 22.5021
\end{array}
\right)$$
