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  1. What is the value of $5^{\log_{5} 2}$?
  2. If $m = y$, which expression is equivalent to $\log (100 m^2$)? Which base logarithm do I use? Is it self implied that its $\log_{10}$? How do I solve it?
  3. When I get a question such as $2\log\frac{m}{n}$ and proceed to simplify it to $\log m - \log n$, do I put the $2$ only on one side or on both? I assume it's only on one side.
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This sounds like homework. What have you done so far? –  mixedmath Aug 28 '11 at 21:22
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For 1, what is the definition of $\log_b(x)$? For 2, I've seen no base specified meaning either a base of 10 or a base of $e$, though it's usually specified somewhere. If these are from a book, does the book say what the generic notation is? I'm also not too sure what the question is asking. $\log(100y^2)$ would be equivalent, but unlikely the answer you desire... For 3, $\log(m/n) = \log(m)-\log(n)$, so multiplying the entire left side by 2 means you want to multiply the entire right side by 2 as well. –  Alex Aug 28 '11 at 21:26
    
In Chrome, if I zoom in to 120%, refresh the page, the $2$ is in the wrong place (it's above the "3.", but under the "self implied that... solve it?" line). Weird. –  muntoo Aug 28 '11 at 21:39

2 Answers 2

up vote 5 down vote accepted

You should read this Dr. Math article.

  1. We start with:
    $$m = 5^{\log_{5} 2}$$ Let's change that to:
    $$m = 5^n, \space\space\space\space\space\space n = \log_{5} 2$$ $n$ represents the number "5 to the power of number is 2". (In other words, $5^n = 2$.) So $m = 5^n$ is the same as $m = 2$.
  2. Usually, mathematicians* mean $\log_{e}$ or $\ln$ when they say $\log$, but that isn't always the case. In your case, I see a $100$, so I assume it's $\log_{10}$.
  3. $2 \cdot \log\frac{m}{n} = 2(\log m - \log n) = 2\log m - 2\log n$.

*In different fields, such as Computer Science, $\log$ can mean even more bizarre things like $\log_{2}$! In finance/economics, it's usually $\log_{e}$.

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Thanks for trying to help me out :). Why is 5^n = 2? –  John Aug 28 '11 at 21:40
    
@John Because that's the definition of a logarithm, if you look in your textbook. Something you might find interesting: the $5$ in $\log_{5}$ is the base; the $5$ in $5^n$ is also the base! When you ask, "What power do I raise 5 to to get 2?", you're really saying: $answer = \log_{5} 2$. –  muntoo Aug 28 '11 at 21:48
    
I recommend reading this Dr. Math article. –  muntoo Aug 28 '11 at 21:51
    
your awesome :). –  John Aug 28 '11 at 21:53
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@John: Yes. If $\log_2(x) = 5$, then $2^5=x$. In general, $\log_a(x)=r$ is equivalent to $a^r=x$. –  Arturo Magidin Aug 28 '11 at 22:03
  1. Remember that the logarithm is the inverse of the exponential. That means that for any base $a$, $$a^{\log_a(x)} = x\text{ for all }x\gt 0;\quad\text{and}\quad \log_a(a^x) = x\text{ for all }x.$$ That is: the logarithm base $a$ "undoes" what the exponential $a^x$ does, and vice versa: just like "adding $1$" and "subtracting $1$" cancel each other (if you first add one, and then you subtract one, it amounts to doing nothing). In the same way, if you first take the logarithm base $a$, and then you take the exponential base $a$ of the answer, it amounts to doing nothing (you get what you started with back); and if you first take the exponential base $a$, and then take the logarithm base $a$ of the answer, then you get back what you started with.

  2. I don't understand what the "$m=y$" is; nor how it relates to the rest of your question. Exactly what $\log$ represents depends on the context. For most people, $\log$ represents the logarithm base $10$, the "common logarithm" (and that is what it means in calculators); in Computer Science, $\log$ often is used to mean $\log_2$ (logarithm base $2$) because this is the most useful logarithm in computer science. In advanced mathematics, $\log$ often represents the natural logarithm (logarithm base $e$).

    At your level, I would be on "logarithm base $10$".

  3. "$\log(m/n)$" is not a question, it's an expression. But notice that "$2\log(m/n)$" means "$2$ multiplying the result of computing $\log(m/n)$". Since $\log(m/n)=\log(m)-\log(n)$, that means that $2\log(m/n)$ is equal to: $$2\log(m/n) = 2 \Bigl( \log(m)-\log(n)\Bigr).$$ So... if you expand this, does the $2$ multiply both summands, or just one? In general, if you have $2(a-b)$, is this equal to $2a-b$ or to $2a-2b$?

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Could you please be so kind and translate your answer of number one into english, im very tired and I find it hard to concentrate. Thanks so much for the otherwise elaborative answer. –  John Aug 28 '11 at 21:30
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@John: I thought it was in English. What do you mean? –  Arturo Magidin Aug 28 '11 at 21:31
    
Can you clarify it a bit more. I dont understand your answer :(. thanks –  John Aug 28 '11 at 21:37
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@John Which part don't you understand? Of course, you can look at my own shining answer below! ;) –  muntoo Aug 28 '11 at 21:45
    
thanks a lot. I think I just need some sleep and then review all of this with a lucid mind :) –  John Aug 28 '11 at 21:56

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