Back-Solving Exponential Eqtns: If e^x <=> ln(x); and Log10(x)<=>10^x , How do I resolve 3

[racookpe1978]

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?

 

[IRstuff]

k308^(1/(b308*m308)) would get you started, I think is what you're asking

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[racookpe1978]

Good idea, I'll try that tomorrow morning. Thank you.

 

[Denial]

I'm a bit confused about what your question actually is, so will concentrate on algebra and "inverting exponential powers".[ ]

If
a = b^c
then (almost by definition)
c = log_b(a)
(the logarithm of a to the base b).

The standard formula for changing the base for a logarithm gives us
log_b(a) = log_u(a) / log_u(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.

 

[dik]

Denial... right on!

Dik

 

[xnuke]

I think the question actually is: given y=xⁿ, with y and n known, find x. IRStuff's answer is correct - x is the nth root of y, or x=y^1/n. As an example, use a calculator to find 10^-2, then raise the answer 0.01 to the power of -1/2 and you get 10.

xnuke
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[racookpe1978]

My test cases on the =LOG(a,b) aren't working yet, but no error message either. I'm going to come back to it.
Could be the carbon-based input device between screen and keyboard.

 

[IRstuff]

I guess you haven't seen "Altered Carbon" on Netflix, or you'd have another descriptor to use: "sleeve." ;-)


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[lilliput1]

LN(a) = LN(K)/(bxM)

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.