log trendline formula
 

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?

--
dr chuck

log trendline formula
 

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

"dr chuck" wrote:

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?

--
dr chuck

log trendline formula
 

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_%28ma...al_constant%29
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.


dr chuck wrote:

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?

--
dr chuck

--

Dave Peterson

log trendline formula
 

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...s/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


--
Regards,
Tom Ogilvy


"Joel" wrote in message
...
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

"dr chuck" wrote:

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?

--
dr chuck

log trendline formula
 

Tom,

You never cease to amaze me with your capabilities.

Alan


"Tom Ogilvy" wrote in message
...
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...s/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


--
Regards,
Tom Ogilvy


"Joel" wrote in message
...
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

"dr chuck" wrote:

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?

--
dr chuck

log trendline formula
 

Alan,

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

--
Regards,
Tom Ogilvy

"Alan" wrote in message
...
Tom,

You never cease to amaze me with your capabilities.

Alan


"Tom Ogilvy" wrote in message
...
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...s/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


--
Regards,
Tom Ogilvy


"Joel" wrote in message
...
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

"dr chuck" wrote:

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?

--
dr chuck