log trendline formula

[Guest]

This is a log trendline formula from Excel semilog graphing.

y = 0.8065Ln(x) + 3.4412

can anyone tell me how the " 0.8065Ln(x) " part of this formula works?

I know y and i am solving for "x"

not sure the significance of the "Ln"

do i have to use some "log" function?

 

[Guest]

ln IS A FUNCTION IN EXCEL

Log is usually referes to base 10
Log(10) = 1
Log(100) = 2
Log( 1000) = 3

If we did Log for base 3
Log(3) = 1
Log(9) = 2
Log(27) = 3

LN is the Log using e as the base where e equals approximately 2.3

 

[Dave Peterson]

ln() is the natural log of a number. It's the exponent that you raise e (2.718
approx) to go get that number.

=ln(2.718) = about 1
since
e^1 is about 2.718

ln(100) = 4.60517
since
e^4.60517 (or 2.718 ^ 4.60517) = 100.

Some info for ln() and exp():
http://en.wikipedia.org/wiki/Natural_logarithm
http://en.wikipedia.org/wiki/E_(mathematical_constant)
http://en.wikipedia.org/wiki/Exponential_function


Starting with:
y = .8065 * ln(x) + 3.4412
subtracting 3.4412 from both sides:

y-3.4412 = .8065 * ln(x)
dividing both sides by .8065

(y-3.4412)/.8065 = ln(x)
Using each side as an exponent with e the base

exp((y-3.4412)/.8065) = exp(ln(x))

But exp(ln(x)) = x

so
exp((y-3.4412)/.8065) = x

So if you put the value for Y in A1, you could use this formula in B1:
=exp((a1-3.4412)/.8065)
to find what X is.

 

[Tom Ogilvy]

LN is called the natural logarithm. It is a logarithm using the exponential
constant (2.718. . .) as the base (see below)

the arguments for log are


LOG(number,base)

Base is optional and if left out, defaults to base 10

so Log(10,10) = 1
Log(10) = 1

but
Log(3) is Log(3,10), not base 3 and equals 0.477121254719662
Log(3,3) = 1 is the logarithm of 3 using base 3.

like wise

EXP(1) gives the exponential consant - it is the exponential constant raised
to the 1st power
http://www.mathcentre.ac.uk/students.php/all_subjects/graphs/exponential_functions/resources/41

=EXP(1) = 2.71828182845905 (not 2.3)

so
=LOG(EXP(1),EXP(1)) is 1
as is
=LN(exp(1))

so LN is equal to LOG with a base of exp(1), or the match constant e

 

[Alan]

Tom,

You never cease to amaze me with your capabilities.

Alan

 

[Tom Ogilvy]

Alan,

Well, Dave Peterson actually gave a more complete response - but thanks!
Hopefully I won't disappoint.