Question related to "Can you help me to solve the below
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
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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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