How do I reverse this formula to give concentration based on the pH

[born2dive00]

Hello everyone I have a formula that will tell me the ph of a liquid based on the concentration.

=-LOG((3)*(1.5*10^-4))
(the 1.5 is the amount of ingredients)

The answer of this formula is a pH of 3.346787486 (and YES I NEED ALL THE NUMBERS)

But How do I change the formula to give me the concentration based on the pH?

So I need a formula that will reverse this so when I enter the ph of 3.346787486 (((((())))))) = 1.5

Please help

Thank you!

Brad

[MrShorty]

The first step in solving this is to go all the way back to beginning algebra when you first learned about logarithms and remember what logarithms mean: https://www.purplemath.com/modules/l...m#Relationship
-LOG(number,10)=pH -> multiply both sides by -1 (remember that pH is a base 10 logarithm and that the LOG() function assumes base 10 unless you specify otherwise).
LOG(number,10)=-pH -> rearrange according to the definition of logarithm and solve for number.

Is that enough of a nudge, or is there something about logarithms that you are having trouble with?

[born2dive00]

I think I did this correct but can some one check my work please. It is imperative that it be super accurate for some calculations I am doing.

The formula I came up with is as follows
=((10^-3.346787486)/3)*10^4
this will equal 1.5 amount of the ingredients

Please let me know Thanks
Brad

[MrShorty]

It looks correct to me. I probably would have stopped with (10^-3.346...)/3 and had the result output as 1.5E-4, but, if your goal is to get just the mantissa only (1.5) of that number, then you would follow by multiplying by 1E4 like you're doing.

[joeu2004]

=-LOG((3)*(1.5*10^-4)) [....] The answer of this formula is a pH of 3.346787486 [....] I need a formula that will reverse this so when I enter the ph of 3.346787486 (((((())))))) = 1.5

First, if 1.5 is a variable ("amount of the ingredient") and 3 is a constant, I would suggest that you write that:

=-LOG(1.5 * 3*10^-4)
or
=-LOG(1.5 * 3E-4)

If the latter formula is in A1, then it becomes clear that the inverse function is:

=10^(-A1)/3E-4

Second, I would not truncate A1 to 3 decimal places (3.346), because the inverse results in 1.50272234847153 instead of 1.5.

Properly rounding A1 to 3 decimal places (3.347) results in 1.49926618297763.

Even rounding to 9 decimal places (3.346787486) results in 1.50000000077593. That might make a difference if you compare with 1.5.

Using the "exact" value in A1 (3.34678748622466 - 3.55E-15) displays 1.500000000000000.

Even that exact result is 1.5 - 1.78E-15, which might make a difference when you compare with 1.5 in some contexts.

Using the displayed value in A1 (3.34678748622466) displays 1.49999999999999.

With all of these calculations, the final result should be explicitly rounded to the accuracy that you require.