How to find the number that the log was derived from.  We will use the bicarbonate system as an example.

If you know the pH of a solution such as urine, blood, intracellular fluid, intercellular fluid, and you know the pKa for a weak acid in that fluid, you can determine the ratio of the [conjugate base] /[weak acid]. First you subtract the pKa from the pH and you get an equation like x=Log [HCO₃^-]/[H₂CO₃].

 0.3 = Log [HCO₃^-]/[H₂CO₃], what is [HCO₃^-]/[H₂CO₃] equal to?

 There are two ways to work this problem.  The first is by remembering the 0.3 is the Log 2 so [HCO₃^-]/[H₂CO₃] = 2

 The second way is to have a calculator with a 10^X button on it.  You simply enter 0.3 and depress the 10^X button.  The answer 2 (or 1.99999) should appear.

Example 2:  If the log of the ratio of the salt to the acid is 1, what is the ratio of the salt to the acid?  That is, if 1.0 = Log [HCO₃^-]/[H₂CO₃], then what is [HCO₃^-]/[H₂CO₃]. 

There are two ways to work this problem.  The first is to remember that the log of 10 is 1.  If that is the case, then [HCO₃^-]/[H₂CO₃] must equal 10.

 The second way is to enter 1 into your calculator and depress the 10^X button.  10 should appear as the answer.

Example 3:  If the log of the ratio of the salt to the acid was –1, what would the ratio of the salt to the acid be.?  That is, if –1 = Log [HCO₃^-]/[H₂CO₃], then what would [HCO₃^-]/[H₂CO₃] be equal to?

 If you do not like working with negative numbers, remember that –1= Log [HCO₃^-]/[H₂CO₃] is the same as 1 = Log [H₂CO₃]/ [HCO₃^-].

 Then there are two ways to work the problem.  First, remember that the log of 10 is 1 so the ratio of  [H₂CO₃]/ [HCO₃^-] must be 10.  By inverting, the ratio of [HCO₃^-]/[H₂CO₃] = 0.1.

 1.6 = Log [HCO₃^-]/[H₂CO₃], what is [HCO₃^-]/[H₂CO₃] equal to?

There are three ways to work this problem.  The first is by changing the 1.6 into 1 + 0.3 + 0.3. Next, remember that the Log of xyz is equal to the Log x + Log y +Log z. So 1 + 0.3 + 0.3 is derived from the Log 10 X 2 X 2 = Log 40. So [HCO₃^-]/[H₂CO₃] = 10 X 2X 2 =40

 The second way is to have calculator with a 10^X button on it.  You simply enter 1.6 and depress the 10^X button.  The answer 40 appears.

A third way is to remember that when the log x changes by 0.3, then x changes by a factor of 2.

 If you had memorized that for normal blood, the values for the Hendersen-Hasselbach equation, pH = pKa + Log [HCO₃^-]/[H₂CO₃], are 7.4 = 6.1 + Log [24mm]/[1.2 mm]. If the pH is increased by 7.3 to 7.6, that is , by 0.3, then the ratio [24mm]/[1.2 mm] had to increase by a factor of 2. that is, it had to increase from 20/1 to 40/1.