how do I compare those two logs? I am having trouble how to show which is larger.
$\log4\cdot \log3 \quad\text{or}\quad \log4 - \log3$
base is $10$ for every $\log$.
I really appreciate your help
 A: Note that:
$$\begin{align}\log4\cdot \log3 \quad&\text{or}\quad \log4 - \log3 \iff \\
\log 3^{\log 4} \quad &\text{or} \quad\log \frac43 \iff \\
3^{\log 4+1} \quad &\text{or} \quad 4 \iff \\
3^{\log 4+1} \quad >3^{1/2+1} =\sqrt{27}\quad > 5 \quad &> \quad 4.\end{align}$$
A: I'm not an expert, but this is what I would do if I want to avoid just plugging it into the calculator. 
I'd try to get everything in terms of something similar, so I'm going to go off of $\log_{10}2$.
$\log 4*\log 3$ is basically $\log 2^2* (\text{something slightly larger than $\log 2$})$. Point is, it's $\log 2$ where the power of 2 is something greater than one. 
$\log 4-\log 3$ is rewritten $\log(4/3)$, and 4/3 is 2 to the power of something less than one. 
I've compared the powers of 2 within the logarithm, and since logs are an increasing function, the one with the bigger power of 2 within is the larger answer. $\log 4*\log 3$ is the larger of the two.
A: On the one hand, $\log 4$ is at least $\frac{1}{2}$, because $4^2=16>10$. Thus the inequality $\log 4 \log 3 \geq \frac{\log 3}{2}$ holds.
On the other hand, $\log 4 - \log 3 = \log (4/3)$, which is less than $\frac{\log 3}{2}$. (Indeed, $(4/3)^2 < 3$ holds, use this to see the claimed inequality).
Putting these together gives $\log 4 \log 3 \geq \frac{\log 3}{2} > \log 4 - \log 3$.
[Meanwhile here is one exercise for the reader: Suppose the base of the logs were 100 or 1000, instead of 10. Would that change things? (Yes it would]
