## 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)

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

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

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

ucan be anything you want (within reason) but would usually beeor10. 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

^{n}, 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

"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand,

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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.