Solving Logarithms (Possibly Natural)?

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Deo3560

New member
Joined
Jul 30, 2010
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Ok, I am kinda new to Logarithms, and they only gave 3 sentences of information about the natural logarithms.

The equation is:

2[sup:1jvh96nl]x[/sup:1jvh96nl]=30

I checked the answer on a online calculator which gave me the answer of:

x=[sup:1jvh96nl]Ln(30)[/sup:1jvh96nl]/[sub:1jvh96nl]Ln(2)[/sub:1jvh96nl] or x?4.90689

I don't really get how 2[sup:1jvh96nl]x[/sup:1jvh96nl]=30 is a natural logarithm
I tried reading other resources that describes natural logarithm, but the only "useful" thing i could find was in Wikipedia.
It said
Wikipedia said:
The natural logarithm is generally written as ln(x), log[sub:1jvh96nl]e[/sub:1jvh96nl](x) or sometimes, if the base of e is implicit, as simply log(x).
Click to expand...

But what I don't understand is, how would I know if "log(x)" is implied to be "Ln(x)", or is it just always that way.

I know how to do the problem with using Log isntead of Ln, but I just don't know if it is the way I should be doing it, incase it comes up on a test or something.

Here's how I did it:

2[sup:1jvh96nl]x[/sup:1jvh96nl]=30
Log(2[sup:1jvh96nl]x[/sup:1jvh96nl])=Log(30)
xLog(2)=Log(30)
.30103x=1.47712
[sup:1jvh96nl].30103x[/sup:1jvh96nl]/[sub:1jvh96nl].30103[/sub:1jvh96nl]=[sup:1jvh96nl]1.47712[/sup:1jvh96nl]/[sub:1jvh96nl].30103[/sub:1jvh96nl]

x?4.90689

I see that the problem does give the same answer, but I just would like to know about the "Ln(x)=Log(x)" question. And if there is a special way that I am supposed to do it by, I would liek to know.

Thanks,
Deo3560
 
I just went over there in my class.
It isn't a natural log per se. What they did there was use the change of base formula

Assuming you know the definition of a log, the problem 2^x = 30 can be rewritten "log base-2 30 = x" I haven't learned how to do the fancy formating some people use in here yet.

Well there isnt a log base-2 button on your calculator you have "log" and "ln". log is log base-10 and ln is log base-e.

To solve this aproxamately you use the change of base formula, which can use log or ln it'll work both ways.

log base-a (b) = log b/log a

so "log 30/log 2 = x" or "ln 30/ln 2 = x", you'll get the same solution either way.
 
Ok thanks,

P.S. : If you want to know how to do the fancy symbols/format:

(For all [----]'s the second one in each pair will have a foreward slash before the word Ex. [word][/word])
(I can't type it in without it turning into the code =P)

Subscript[sub:155wksm9]Subscript[/sub:155wksm9] : [sub:155wksm9]Subscript[/sub:155wksm9] (Note: Subscripts may be cut off if you do not have another line of text somewhere below it)

Superscript[sup:155wksm9]Superscript[/sup:155wksm9] : [sup:155wksm9]Superscript[/sup:155wksm9]

Fraction [sup:155wksm9]Superscript[/sup:155wksm9]/[sub:155wksm9]Subscript[/sub:155wksm9] (Note: the / is not in any BBCode.)


Also there are other Characters such as:
? ? ? ? ? ± ? ? ÷ ? ? ?

In Windows, you can click your "Start" button and either search for "Character Map"
or go to:
C:\ProgramData\Microsoft\Windows\Start Menu\Programs\Accessories\System Tools
 
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