Back-Solving Exponential Eqtns: If e^x <=> ln(x); and Log10(x)<=>10^x , How do I resolve
Back-Solving Exponential Eqtns: If e^x <=> ln(x); and Log10(x)<=>10^x , How do I resolve
(OP)
It might be my rusty algebra, or my even slower Excel, but I have a question:
If cell [g308] "= exp^(f308)", then its inverse (in Excel) in cell [h308] is "= LN(g308)"
and cell [f308] is going to equal cell [h308]
If cell [p308] "= Log(q308)", then its inverse in cell [r308] "= 10^(p308)"
and like above, cell [p308] will equal cell [r308]
But what is the most efficient way to invert other exponential powers?
Algebraically, I had been solving for K from K = a^(b*M) for known values of a, b, and M.
My original equation in cell [k308] was "= a308^(b308*M308)"
Works fine against the master reference.
Experimental results changed, now I have to check for variations of "a" as a function of b, M, and K.
What are my most efficient equation in cell [a308] to invert that formula?
If cell [g308] "= exp^(f308)", then its inverse (in Excel) in cell [h308] is "= LN(g308)"
and cell [f308] is going to equal cell [h308]
If cell [p308] "= Log(q308)", then its inverse in cell [r308] "= 10^(p308)"
and like above, cell [p308] will equal cell [r308]
But what is the most efficient way to invert other exponential powers?
Algebraically, I had been solving for K from K = a^(b*M) for known values of a, b, and M.
My original equation in cell [k308] was "= a308^(b308*M308)"
Works fine against the master reference.
Experimental results changed, now I have to check for variations of "a" as a function of b, M, and K.
What are my most efficient equation in cell [a308] to invert that formula?





RE: Back-Solving Exponential Eqtns: If e^x <=> ln(x); and Log10(x)<=>10^x , How do I resolve
TTFN (ta ta for now)
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RE: Back-Solving Exponential Eqtns: If e^x <=> ln(x); and Log10(x)<=>10^x , How do I resolve
RE: Back-Solving Exponential Eqtns: If e^x <=> ln(x); and Log10(x)<=>10^x , How do I resolve
If
a = bc
then (almost by definition)
c = logb(a)
(the logarithm of a to the base b).
The standard formula for changing the base for a logarithm gives us
logb(a) = logu(a) / logu(b)
where u can be anything you want (within reason) but would usually be e or 10. However this step is not required within Excel because its LOG() function has an optional second argument that is the base to which the logarithm is to be taken. ; Thus you get the required result with
=LOG(a,b)
HTH.
RE: Back-Solving Exponential Eqtns: If e^x <=> ln(x); and Log10(x)<=>10^x , How do I resolve
Dik
RE: Back-Solving Exponential Eqtns: If e^x <=> ln(x); and Log10(x)<=>10^x , How do I resolve
xnuke
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RE: Back-Solving Exponential Eqtns: If e^x <=> ln(x); and Log10(x)<=>10^x , How do I resolve
Could be the carbon-based input device between screen and keyboard.
RE: Back-Solving Exponential Eqtns: If e^x <=> ln(x); and Log10(x)<=>10^x , How do I resolve
TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm
RE: Back-Solving Exponential Eqtns: If e^x <=> ln(x); and Log10(x)<=>10^x , How do I resolve
a = Exp((LN(K)/(bxM))
but you have to correlate with your data at various values of K to confirm validity of the equation for K. You may have to define range of K for particular value of a, b & M to be valid.