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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- July 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?

- July 26th 2007, 02:28 PMCaptainBlack
- July 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. - July 26th 2007, 03:53 PMPlato
Your problem may well be in notation. The way you write the problem.

Here is an example.

- July 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. - July 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 , which is a matter of notation, as Plato suggested.

Your calculator doesn't have a log button that will do .

What you did was

So let's rethink your original problem. Set

According to Plato's post this is equivalent to:

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

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 - July 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. - July 27th 2007, 11:17 AMtopsquark
Ah, my apologies. I had misunderstood the post.

We commonly write just "log" when we mean .

So your expression is using two different conventions, which is why you are confused. and 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 is the same as where . All I can say is that conventions vary and that your calculator almost certainly uses the convention that .)

-Dan