Solve the following inequalities

Question

Does there exist a general formula for comparing two prime numbers powers . Ex. $$ max(2^{2002},3^{1335} )$$ or $$ max(2^{2004} ,3^{1202} ) $$ NOTE: Im a twelfth grader and i accidentally stumbled across these problems.I am curious for an universal method or ,at least , for their solution.

Answer 1

I'll explain a general method (will not work for all problems but for your problems is enough ) . This is essentially equivalent with the logarithm method in some way .

First find some $x$ and $y$ such that $3^x >2^y$ . Use fairly big values and not something like $3^2>2^3$ . For example $3^7=2187>2048=2^{11}$ would suffice .

Now suppose you want to prove something like $3^a>2^b$ .Use our inequality $3^7>2^{11}$ and raise it to the power $\frac{a}{7}$ so :

$$3^a>2^{\frac{11a}{7}}$$

Now with a little wishful thinking we would want that $\frac{11a}{7}\geq b$ and then we would be done :

$$3^a>2^{\frac{11a}{7}}\geq 2^b$$ as wanted .So we want that $\frac{a}{b}\geq \frac{7}{11}$

Note :This is where the similarity with the logarithm method is clearly seen :

Using the logarithm method we would compare $\frac{a}{b}$ with $\frac{\log 2}{\log 3}$ but now we compare it with our approximation for $\frac{\log 2}{\log 3}$ namely $\frac{7}{11}$ .

In this case we're lucky because it can be verified that $\frac{a}{b}=\frac{1335}{2002}>\frac{7}{11}$ (use WolframAlpha if you're lazy :))

Hope you can now solve the other problem . Good luck .