I have Tf1(ADC1), but need to find out ADC1
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Is Tf1 a function and ACD1 a variable? – mrnovice Mar 29 '17 at 04:41
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@mrnovice yes. So in the above equation if i assume ADC = 1, i'll get Tf1(ADC1) as -98. All good. Now what i need is when i have -98 with me and don't have the value of ADC. – Robin C Samuel Mar 29 '17 at 05:34
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ok, well have a look at the answer I wrote – mrnovice Mar 29 '17 at 05:35
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Let $f = Tf_{1}$, let $x = ACD_{1}$
We have $$f(x) = \frac{9}{\frac{\ln(\frac{c}{x}-\frac{1}{5})}{676}+d}-459.67$$
with $c = 207.08502024291497975584, d= 0.016770082173402649673$
$$(\frac{\ln(\frac{c}{x}-\frac{1}{5})}{676}+d)f(x) = 9 - 459.67(\frac{\ln(\frac{c}{x}-\frac{1}{5})}{676}+d)$$
$$(f(x)+459.67)(\frac{\ln(\frac{c}{x}-\frac{1}{5})}{676})= 9 -459.67d -df(x)$$
$$\frac{\ln(\frac{c}{x}-\frac{1}{5})}{676} = \frac{ 9 -459.67d -df(x)}{f(x)+459.67}$$
$$\frac{c}{x}-\frac{1}{5} = e^{ \frac{676( 9 -459.67d -df(x))}{f(x)+459.67}}$$
$$\frac{c}{x} = \frac{5e^{ \frac{676( 9 -459.67d -df(x))}{f(x)+459.67}}+1}{5}$$
$$x = \frac{5c}{5e^{ \frac{676( 9 -459.67d -df(x))}{f(x)+459.67}}+1}$$
mrnovice
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@RobinCSamuel I don't understand what you mean - what exactly do you want? – mrnovice Mar 29 '17 at 06:01
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