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

  • $\begingroup$ Is there anything wrong with using a calculator? $\endgroup$ – Yonatan N May 31 '18 at 21:43

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]

  • $\begingroup$ would I need a calculator if the base were 100? $\endgroup$ – Hirotaka Nakagame Jun 4 '18 at 18:36

Note that: $$\begin{align}\log4\cdot \log3 \quad&\text{or}\quad \log4 - \log3 \iff \\ \log 3^{\log 4} \quad &\text{or} \quad\log \frac43 \iff \\ 3^{\log 4+1} \quad &\text{or} \quad 4 \iff \\ 3^{\log 4+1} \quad >3^{1/2+1} =\sqrt{27}\quad > 5 \quad &> \quad 4.\end{align}$$

  • 3
    $\begingroup$ I think your chain would be more clear if you added the intermediate step between $\log 3^{\log 4} \text{or} \log \frac{4}{3}$ and $3^{\log 4 + 1} \text{or} 4$. $\endgroup$ – NoOneIsHere May 31 '18 at 20:39
  • 1
    $\begingroup$ or get the log 3 to the other side on the first step and factor it out. $\endgroup$ – WorldSEnder Jun 1 '18 at 1:01
  • $\begingroup$ @NoOneIsHere, good point, but dropping logs seemed too easy step, so I multiplied both sides by 3 at once. Thank you for commenting. $\endgroup$ – farruhota Jun 1 '18 at 3:30
  • $\begingroup$ @WorldSEnder, good point, though it is a little intricate. I tried to get rid of one of the logs as fast as possible. Thank you for commenting. $\endgroup$ – farruhota Jun 1 '18 at 3:31

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.

  • 1
    $\begingroup$ Is there any particular reason for all the \mathsfs everywhere? You can simply use \log for log $\endgroup$ – Au101 May 31 '18 at 21:27
  • $\begingroup$ @Au101 Nope! Thanks for letting me know - I'm new at this and didn't know you could use \log instead. $\endgroup$ – Eri May 31 '18 at 21:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.