I dont understand how log(2)(2) = 1. Inserting it in my calculator returns ~.6 but my textbooks says it equals 1. Is it just some sort of fact I have to believe, or is there a way to determine it?

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- Jul 26th 2007, 02:18 PMceasar_19134Logarithmic Functions
I dont understand how log(2)(2) = 1. Inserting it in my calculator returns ~.6 but my textbooks says it equals 1. Is it just some sort of fact I have to believe, or is there a way to determine it?

- Jul 26th 2007, 02:28 PMCaptainBlack
The definition of $\displaystyle \log$ to the base $\displaystyle b$ of $\displaystyle a$ is the power that $\displaystyle b$ must be raised to

to give $\displaystyle a$. So as $\displaystyle b^1=b$, we have $\displaystyle \log_b(b)=1$ for all positive bases $\displaystyle b$.

RonL - Jul 26th 2007, 02:40 PMceasar_19134
Im still having trouble understanding. Perhaps it would simply be easier to explain how to go about solving log(2)(2) or log(2)(3)?

The book doesnt explain at all- It simply shows of table of values when x = 2, 4, 8, 16, ..., 128. - Jul 26th 2007, 03:53 PMPlato
Your problem may well be in notation. The way you write the problem.

Here is an example.

$\displaystyle \begin{array}{l} Log_b \left( a \right) = x\quad \Leftrightarrow \quad b^x = a \\ Log_3 \left( {81} \right) = 4\quad \Leftrightarrow \quad 3^4 = 81 \\ \end{array}$ - Jul 27th 2007, 10:26 AMceasar_19134
Didnt get very far before I got stuck again-

"The base-10 logarithm is called the common logarithm. The common logarithm, log(10)x, is usually written as log x."

I dont understand what happens to the 10. The equation could be arranged into 10^y=x, but that doesnt help me any, if anything its causes more confusion. - Jul 27th 2007, 10:49 AMtopsquark
I have an idea of what happened initially anyway.

You plugged log(2)(2) into your calculator and got 0.60206, didn't you?

The question your book is asking about is $\displaystyle log_2(2)$, which is a matter of notation, as Plato suggested.

Your calculator doesn't have a log button that will do $\displaystyle log_2$.

What you did was

$\displaystyle log (2)(2) = log_{10}(2 \cdot 2) = log_{10}(4) \approx 0.60206$

So let's rethink your original problem. Set

$\displaystyle x = log_2(2)$

According to Plato's post this is equivalent to:

$\displaystyle 2^x = 2$

What x value(s) satisfy this equation? The one that should immediately come to mind is x = 1. And that's the only solution, in fact.

Thus

$\displaystyle x = log_2(2) = 1$

There is a way to get the calculator to work out your problem. I'm not going to tell you explicitly for fear of confusing you more than you already are. The name of the equation is the "Change of base formula." Look it up when you get a little more confidence using logarithms.

-Dan - Jul 27th 2007, 10:58 AMceasar_19134
Perhaps I explained it wrong or am still managing not to see what is under my nose, but I now understand what I had done wrong with log"(2)"2.

However, when presented with log"(10)"x = log x I cant advance- Rearranged yields 10^y=x, which doesnt make any sense to me. - Jul 27th 2007, 11:17 AMtopsquark
Ah, my apologies. I had misunderstood the post.

We commonly write just "log" when we mean $\displaystyle log_{10}$.

So your expression is using two different conventions, which is why you are confused. $\displaystyle log(x)$ and $\displaystyle log_{10}(x)$ are meant to be the same thing.

(Warning: There are a number of members on the site, especially the European ones I think, that will tell you that $\displaystyle log(x)$ is the same as $\displaystyle log_e(x)$ where $\displaystyle e = 2.71828...$. All I can say is that conventions vary and that your calculator almost certainly uses the convention that $\displaystyle log(x) = log_{10}(x)$.)

-Dan