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# Back-Solving Exponential Eqtns: If e^x <=> ln(x); and Log10(x)<=>10^x , How do I resolve3

## 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?

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

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

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

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

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

Denial... right on!

Dik

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

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=y1/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
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
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### RE: Back-Solving Exponential Eqtns: If e^x <=> ln(x); and Log10(x)<=>10^x , How do I resolve

(OP)
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.

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

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

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
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### RE: Back-Solving Exponential Eqtns: If e^x <=> ln(x); and Log10(x)<=>10^x , How do I resolve

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.

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