Alternate soultion for solving an inequality The following image contains both the question and its solution.

In the highlighted section of the solution, we are supposed to solve the inequality to arrive at the final answer. I solved by finding different powers of $\left(-\frac{3}{4}\right)$. But, I don't wish to do the same in exams since it is actually time consuming process. I want to know whether there is any other alternate approach in solving the inequality(except using logarithms, since log tables are not allowed in examination) or any other alternate solution for the entire problem itself. 
 A: The comment above can be an answer, I think : 

No, I think that while you find it time consuming, you are also expected to be "sufficiently practiced" enough to be dealing with this part of the question without creating a time constraint. There is no other way of doing the question : in fact , the construction of the question clearly indicates which formulas to use.

Just see the construction with your eyes : There is every bit the hint that you are to simplify the expression for $a_n$ using the geometric sum formulas you should know. Also , $b_n$ is just a placeholder for $a_n < 0.5$, so it does not play any role in terms of a "different"  approach. There is no alternate approach to the question, your approach is the correct one.
Let us still look at the last line more carefully. Note that the inequality is satisfied if $n$ is even, so we may focus on $n$ odd, say $n = 2k+1$.  In that case, the RHS is $\frac{3^{2k+1}}{4^{2k+1}}$, so we are looking for when $4^{2k+1} > 6 \times 3^{2k+1}$. After some power-related simplification, this comes down to : find $k$ for which $2 \times 16^{k} > 9^{k+1}$.
And that is it : now you try values of $k$. I am sure $k=0$ and $k=1$ don't work,  that is easy to see. It is not so difficult to see that $k=2$ doesn't work, both sides are three digit numbers. That $k=3$ works is finding four digit numbers on either side for comparison.  
So the answer is $6$, the even number before $2 \times 3 + 1$.

It can be made out very clearly that your exam is from an Indian context. I probably gave the exams as you at your age,  and I went through some absolutely shambolic things just like you : in the name of mathematics, they sell computation. For example, there is no simplification of the integral $\int \sqrt{x^2 - a^2}$ : it just is as messy as it looks, drink it in.
To force myself through this, I actually had to end up memorizing power values of small numbers.

In particular, memorize powers of $2,3,5,7$ up till the powers are five digit numbers(What of it, I went up till $2^{32} = 4294967296$. To keep memorization in place, I set them(and other formulas) as passwords on some sites, so  I would need to recall them again and again.).

Spend some time doing this, and GP questions like this will fall in place.
