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David Biddulph[_2_] David Biddulph[_2_] is offline
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Default Question related to "Can you help me to solve the below

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


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