When you punch in log on your calculator

[DB]

b^x=y
log_b(y)=x

When you punch in log on your calculator, mathematically, how is it solving for x?

For example,
3^x=81
log_3(81)=4
How is this being solved?

Thanks

 

[Jameson]

I don't have a calculator where you can define the base like that.

You rewrote the original question that skipped a lot of steps.

I would do the question like this:

3^x = 81
x log(3) = log(81)
x = \frac{log(81)}{log(3)}

I don't know exactly how the calculator evalulates the value, but those are the steps you can take to see a process.

 

[Crosson]

I used to ask this question, the answer given to me was "the calculator memorizes the values", this seemed reasonable, but lame.

The truth is that logarithms, sines, cosines, anything, can be expanded as an infinite polynomial. For example:

ln(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 + ...

sin (x) = x - (x^3)/3! + (x^5)/5! - ...

Where ln is log base e and 5! means 5*4*3*2*1. These series have an infinite number of terms, so the calculator has to cut them off at some point (approximate).

 

[Chrono]

ln(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 + ...

It has -1 < x < 1 after the expansion in a reference book of mine.

And, yeah, expansions is what caluculators use to find values of a given x.

 

[damoclark]

ln(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 + ...

Generally , this series converges too slowly to be of any use to calculate ln(1+x). Consider if x=100, you would need to calculate zillions of terms in the series for it to be of any use, I'm not even sure it would converge for x>1. Of course if 0<x<1, the series converges quite fast.

Consider the problem of calculating ln(1000), this is how I think your calculator does it.

ln( (1+x)/(1-x) ) = ln(1+x) - ln(1-x)

Set (1+x)/(1-x) = 1000 => x = 999/1001


therefor ln(1+999/1001) - ln(1-999/1001) = log(1000)

Now since you value of x is less than one, your calculator doen't have to use so many terms in the series to calculate an accurate answer, as convergence is acheived quite quickly.

 

[DB]

Sorry, but wats ln?

 

[DeadWolfe]

ln = log to the base e

It's often just denoted log.

 

[Chrono]

It's often just denoted log.

DB, I suggest you memorize this.

 

[DB]

lol, kk thanks. I know e is approx = 2.17... so I am going to study the posts and see If I can understand, I'll probably have more questions.

 

[Chrono]

lol, kk thanks. I know e is approx = 2.17... so I am going to study the posts and see If I can understand, I'll probably have more questions.

The reason I said that is because in calculus you will see ln denoted as log and I think it's pretty safe to say that it's generally accepted like that (I still remember that thread where we had a lengthy discussion about this).

And we're always here whenever you want to ask more questions.

 

[dextercioby]

I frankly doubt it.I think it should be "ln" everywhere...What about logarithm to the base of 10,how would you write that ...?

Daniel.

P.S.BTW:\ln voilà...

 

[Oggy]

e=2.7182, not 2.17.

 

[dextercioby]

Not really,it's an transcendental irrational #,therefore
e\approx 2.7183

Daniel.

 

[Oggy]

Yeah of course, and now I know the Latex code for \approx :)

 

[Chrono]

I frankly doubt it.I think it should be "ln" everywhere...What about logarithm to the base of 10,how would you write that ...?

Maybe like this:

\log_{10}

 

[dextercioby]

Ha,ha,why that way,when log would mean automatically base "e"...?;bugeye:

Daniel.

 

[Oggy]

I thought \log_{e}{x}=\ln{x}, and \log_{10}{x}=log{x}.

 

[dextercioby]

It would be a good option,however,we in Romania used the best:
\log would mean any base,except "e" & 10.You would have to specify the base as a subscript.E.g. \log_{8\sqrt{7\pi}}
\ln [/itex] would mean base &quot;e&quot;.<br /> \lg [/itex] would mean base 10...&lt;br /&gt; &lt;br /&gt; Daniel.

 

[dextercioby]

What do you know,LATEX recognizes all three notations,hopefully with the definitions that i specified...

Daniel.

 

[houserichichi]

I'd actually learned it as log being base 10, ln being base e, and lg being base 2...for all those wacky computer scientists, no less.

 

[dextercioby]

I wasn't any "wacky computer scientist",though.I liked this part of mathematics,anyway...

Daniel.

 

[Chrono]

\lg [/itex] would mean base 10...

<br /> <br /> I&#039;ve never seen that before. <br /> <br /> Well, all I know is that all my calculus professors told us that \log means base e and if you wanted it to mean anything else you would have to specify the base.<br /> <br /> On a side note, Mathematica, the program we all know and love, does it that way, too. From it&#039;s help browser it states: &quot;Log[z] gives the natural logarithm of z (logarithm to base e)&quot;.

 

[DeadWolfe]

I thought \log_{e}{x}=\ln{x}, and \log_{10}{x}=log{x}.

That is what is taught in high schools.

Among certain groups of people, it is the standard.

Amongst mathematicians, it is not. They use log to indicate a base of e, since it's the log they use the most often. (By FAR! Really. Nothing else really is deserving of comparison.)

 

[Chrono]

That is what is taught in high schools.

Among certain groups of people, it is the standard.

Amongst mathematicians, it is not. They use log to indicate a base of e, since it's the log they use the most often. (By FAR! Really. Nothing else really is deserving of comparison.)

I didn't think I was the only one who thought this was the case.

 

[dextercioby]

\ln makes more sense,because it comes from 'logaritmus naturalis' which automatically specifies the base...
As for mathematicians' notations,well,their definitions are unique,but their notations may vary...

Daniel.

 

[Oggy]

Of course the natural logarithm is used most often, and \ln the notation I like most for it.

 

[CrankFan]

When you punch in log on your calculator, mathematically, how is it solving for x?Thanks

Do a web search on "CORDIC."