Without more information, I believe that they are reporting that 7.1 is the log (base 10) of cfu/ml? If so, then you can convert to cfu/ml = 10^7.1 or about 1.2589 * 10^7 colony forming units per milliliter. If you know the volume and any dilutions they used, you can then back calculate the total number of bacterial cells in the whole culture. Good luck

Without more information, I believe that they are reporting that 7.1 is the log (base 10) of cfu/ml? If so, then you can convert to cfu/ml = 10^7.1 or about 1.2589 * 10^7 colony forming units per milliliter. If you know the volume and any dilutions they used, you can then back calculate the total number of bacterial cells in the whole culture. Good luck

If we have 7.1log no. of colony forming units in a given sample it means we take antilog of it. Seven is Equal to ten million cells and let.0.1indicate2. 1 that means it is eqal to 21million cdlls.

Hi

Please look in attach file

thank you for your answer @ Stewart Gardner . Here is more detail about my question . Simply, it is an experiment to compare the number of bacteria in a dental root canal without dressing the canal with medication and after dressing thevdental root canal with medication . The attached file clarifies what i am saying. The problem that the bacterial account is mentioned in log of cfu so i cant know the real number of bacteria . @stewart garden

@ Dharam Vir Pathak . i have added two attached files. does that change anything related the answer . thank you for you answer :)

Hi,

I think you need to subtract the initial numbers i.e., what was the initial load with which they have started the experiment and see how much was reduced after medication. This will help if you are comparing cfu reduction in with and without medication. If the question is only regarding numbers then answers from @Stewart Gardner, @Dharam vir Pathak holds good. Cheers!!!!!

Amer al shalab,

Based on the documents you added, I interpret the experiment as follows:

They collected some initial sample from the teeth and then diluted this sample 1:10,000 and plated 0.01ml to count CFU (colony forming units) with the standard assumption that 1 cfu = 1 viable bacterial cell. So, if the log of CFU is reported as 1, that means that CFU=10^1=10 colonies on a plate of 0.01ml of 1:10,000 dilution. So, then you convert to 10cfu * 100 = 1.0x10^3 cfu/ml in the 0.01ml sample plated and convert to 1.0x10^3cfu/ml * 10,000 dilution factor = 1.0x10^7 cfu/ml in the initial sample. Now to determine how many bacteria are in the original sample, you need to know the initial volume of the sample. For example if the sample is only 0.5ml, then even though you counted 1.0x10^7cfu/ml, the original sample only has 5x10^6cfu because it was only a half of a milliliter.

Looking at the actual data, let’s use data for Layer 1 after 10 min. treatment, you have 1.27 logCFU for IKI, 4.25 for CH, and 6.73 for the control. So, converting those to actual CFU would give you 18.6, 1.78x10^4, and 5.37x10^6 respectively. Then, converting to cfu/ml (adjusting for the dilutions) would give you 1.86x10^7, 1.78x10^10, and 5.37x10^12 cfu/ml. This seems to show that the IKI has the least bacteria and that the control has the most, which is expected if the control is no treatment. I hope this helps.

I agree with all answers above