What is the difference in this question between $\log$ and $\lg$?

Question

Am I right in assuming that $\lg$ just refers to $\log$ base ($10$)? Whereas $\log$ is just any unspecified log?

I'm solving $\lg{15}-\lg{5}$

Am I good to just use the standard rules of logarithms, where subtraction is concerned?

@NewZealand'sfinest: Yes, except that when you use $\lg$, you don’t add the subscript $2$: it’s understood. You have $\lg 15-\lg 5=\log_215-\log_25$. — Brian M. Scott, May 10 '16 at 21:54
Great, thanks for your help!
I'm happy for this question to be closed if it isn't going to contribute to anybody else. — New Zealand's finest, May 10 '16 at 21:55
Wikipedia suggests lg is $\log_{10}$ while ln is $\log_e$ and lb is $\log_2$, as suggested by ISO 31-11, though some use lg for $\log_2$ — Henry, May 10 '16 at 22:00
@NewZealand'sfinest It would be $2\log_e8$ where $e$ is Euler's constant. — Simply Beautiful Art, May 10 '16 at 22:02
If I were asked to simplify 2ln8 - ln5 + 2ln10, am I able to solve this as before using the log rules and without being concerned by Euler's constant at all? Purely because it states to simplify, and not to solve. — New Zealand's finest, May 10 '16 at 22:21

score 3 · Answer 1 · edited May 17 '16 at 09:14

3

$\lg3$ is the answer :)

Because $\lg{\dfrac{15}{5}}=\lg{3}$

edited May 17 '16 at 09:14

Roby5

4,287

answered May 10 '16 at 22:00

New Zealand's finest

353

If the answer is $lg3$ then you don't really care if it means $\log_2$ or $\log_{10}$ :) – peter.petrov May 17 '16 at 09:40

score 0 · Answer 2 · answered May 17 '16 at 09:38

In general, the use of $lg$ is ambiguous. Therefore this problem is not properly stated, it should have been explicitly mentioned what $lg$ means. Of course, if that's just an exercise after some lesson, and that lesson defines what exactly $lg$ is, then the problem is stated fine.

What is the difference in this question between $\log$ and $\lg$?

2 Answers2