Thanks guys.

A quick background. I get a trendline and select the exponential function.
It comes up with a formular i.e.
y = 3949.4 e "superscript" 0.012x
I have the y number = 10,000
*(actually I also have the first set of number of x = 80.40, my first
conversion is 3949.4*EXP(0.012*80.49)

I used your formular
Log (10,000/3949.4)/0.012 and get the number of 35.08, however, I understand
I shall get a number close to 80. Am I putting the wrong formular?

Thanks
--
--


"joeu2004" wrote:

On Nov 29, 6:27 am, K Lee wrote:
Can you help me to solve the below equation in excel ?
Y= 3494e0.012x and Y=10,000, therefore,
10000=3949e0.012x,
I like to find out the value of x

I wonder if "e" is the mathematical constant (aka Euler's number), and
you meant to write:

Y = 3494*e^(0.012*x)

In that case, we really should require that you complete the algebraic
transformation yourself, but what the heck ....

ln(Y / 3494) = 0.012 * x

x = ln(Y / 3494) / 0.012

In Excel, if Y (10000) is in A1, in A2 you would write the formula:

=ln(A1 / 3494) / 0.012

As a double-check, in A3 you might write:

=3494 * exp(1) ^ (0.012 * A2)

We expect __about__ 10000. The result might not be exact because of
the many approximations and the artifacts of binary computer
arithmetic.

--
--

He didn't say LOG, he said LN.
If you don't know the difference, look them up in Excel help.
--
David Biddulph

"K Lee" wrote in message
...
Thanks guys.

A quick background. I get a trendline and select the exponential function.
It comes up with a formular i.e.
y = 3949.4 e "superscript" 0.012x
I have the y number = 10,000
*(actually I also have the first set of number of x = 80.40, my first
conversion is 3949.4*EXP(0.012*80.49)

I used your formular
Log (10,000/3949.4)/0.012 and get the number of 35.08, however, I
understand
I shall get a number close to 80. Am I putting the wrong formular?

Thanks
--
--


"joeu2004" wrote:

On Nov 29, 6:27 am, K Lee wrote:
Can you help me to solve the below equation in excel ?
Y= 3494e0.012x and Y=10,000, therefore,
10000=3949e0.012x,
I like to find out the value of x

I wonder if "e" is the mathematical constant (aka Euler's number), and
you meant to write:

Y = 3494*e^(0.012*x)

In that case, we really should require that you complete the algebraic
transformation yourself, but what the heck ....

ln(Y / 3494) = 0.012 * x

x = ln(Y / 3494) / 0.012

In Excel, if Y (10000) is in A1, in A2 you would write the formula:

=ln(A1 / 3494) / 0.012

As a double-check, in A3 you might write:

=3494 * exp(1) ^ (0.012 * A2)

We expect __about__ 10000. The result might not be exact because of
the many approximations and the artifacts of binary computer
arithmetic.

--
--

I guess answer should be 92.6819

y = 3949.4 exp(0.012x)
y = 10000

10000 = 3949.4 exp(0.012x)
taking log both sides

log(10000) = 0.012x*log(3949.4)

4 = 0.012x*3.596531

0.012x = 1.112183

x = 1.112183 / 0.012

x = 92.6819

u can use this

=(LOG(10000)/LOG(3949.4))/0.012



K Lee wrote:

Thanks guys.

A quick background. I get a trendline and select the exponential function.
It comes up with a formular i.e.
y = 3949.4 e "superscript" 0.012x
I have the y number = 10,000
*(actually I also have the first set of number of x = 80.40, my first
conversion is 3949.4*EXP(0.012*80.49)

I used your formular
Log (10,000/3949.4)/0.012 and get the number of 35.08, however, I understand
I shall get a number close to 80. Am I putting the wrong formular?

Thanks
--
--


"joeu2004" wrote:

On Nov 29, 6:27 am, K Lee wrote:
Can you help me to solve the below equation in excel ?
Y= 3494e0.012x and Y=10,000, therefore,
10000=3949e0.012x,
I like to find out the value of x

I wonder if "e" is the mathematical constant (aka Euler's number), and
you meant to write:

Y = 3494*e^(0.012*x)

In that case, we really should require that you complete the algebraic
transformation yourself, but what the heck ....

ln(Y / 3494) = 0.012 * x

x = ln(Y / 3494) / 0.012

In Excel, if Y (10000) is in A1, in A2 you would write the formula:

=ln(A1 / 3494) / 0.012

As a double-check, in A3 you might write:

=3494 * exp(1) ^ (0.012 * A2)

We expect __about__ 10000. The result might not be exact because of
the many approximations and the artifacts of binary computer
arithmetic.

--
--

I think you would want to use LN instead of LOG (given the the value being
raised to the given power is 'e' and not 10).

--
Rick (MVP - Excel)


"muddan madhu" wrote in message
...
I guess answer should be 92.6819

y = 3949.4 exp(0.012x)
y = 10000

10000 = 3949.4 exp(0.012x)
taking log both sides

log(10000) = 0.012x*log(3949.4)

4 = 0.012x*3.596531

0.012x = 1.112183

x = 1.112183 / 0.012

x = 92.6819

u can use this

=(LOG(10000)/LOG(3949.4))/0.012



K Lee wrote:

Thanks guys.

A quick background. I get a trendline and select the exponential
function.
It comes up with a formular i.e.
y = 3949.4 e "superscript" 0.012x
I have the y number = 10,000
*(actually I also have the first set of number of x = 80.40, my first
conversion is 3949.4*EXP(0.012*80.49)

I used your formular
Log (10,000/3949.4)/0.012 and get the number of 35.08, however, I
understand
I shall get a number close to 80. Am I putting the wrong formular?

Thanks
--
--


"joeu2004" wrote:

On Nov 29, 6:27 am, K Lee wrote:
Can you help me to solve the below equation in excel ?
Y= 3494e0.012x and Y=10,000, therefore,
10000=3949e0.012x,
I like to find out the value of x

I wonder if "e" is the mathematical constant (aka Euler's number), and
you meant to write:

Y = 3494*e^(0.012*x)

In that case, we really should require that you complete the algebraic
transformation yourself, but what the heck ....

ln(Y / 3494) = 0.012 * x

x = ln(Y / 3494) / 0.012

In Excel, if Y (10000) is in A1, in A2 you would write the formula:

=ln(A1 / 3494) / 0.012

As a double-check, in A3 you might write:

=3494 * exp(1) ^ (0.012 * A2)

We expect __about__ 10000. The result might not be exact because of
the many approximations and the artifacts of binary computer
arithmetic.

--
--