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?
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.
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.
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.
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.
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 .)