removal and inactivation requirements for Giardia in a filtration plant Given initial pathogen concentration (N_0) = 42,860 org/L; final pathogen concentration (N) = 4.4290 org/L, determine the log removal/inactivation credit (logR) in log10.

Start with the thinking

• The governing relation printed in this handbook section is Removal and Inactivation Requirements.
• Everything except logR is given, so isolate logR symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Removal and inactivation requirements set the required log reduction of pathogens between raw and finished drinking water.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\log_{10} R = \log_{10}\left(\dfrac{N_0}{N}\right)$$

2. Step 2 — Rearrange symbolically for logR:

   $$logR = \log_{10}\left(\dfrac{N_0}{N}\right)$$

3. Step 3 — List the givens: initial pathogen concentration (N_0) = 42,860 org/L, final pathogen concentration (N) = 4.4290 org/L.

4. Step 4 — Substitute the given values:

   $$logR = \log_{10}\left(\dfrac{N_0}{4.4290}\right)$$

5. Step 5 — Evaluate:

   $$logR = 3.9857\ \text{log10}$$

6. Step 6 — Check: returning logR = 3.9857 log10 to

   $$\log_{10} R = \log_{10}\left(\dfrac{N_0}{N}\right)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 7.9715 — kept a factor of two that cancels in the correct rearrangement.
• 1.9929 — dropped that same factor in the other direction.
• 4.3843 — rounded an intermediate value before the final step.

Reference: FE Handbook — Removal and Inactivation Requirements

removal and inactivation requirements log credit calculation Given initial pathogen concentration (N_0) = 5,020 org/L; log removal/inactivation credit (logR) = 5.9700 log10, determine the final pathogen concentration (N) in org/L.

Start with the thinking

• The governing relation printed in this handbook section is Removal and Inactivation Requirements.
• Everything except N is given, so isolate N symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Removal and inactivation requirements set the required log reduction of pathogens between raw and finished drinking water.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\log_{10} R = \log_{10}\left(\dfrac{N_0}{N}\right)$$

2. Step 2 — Rearrange symbolically for N:

   $$N = \dfrac{N_0}{10^{\log R}}$$

3. Step 3 — List the givens: initial pathogen concentration (N_0) = 5,020 org/L, log removal/inactivation credit (logR) = 5.9700 log10.

4. Step 4 — Substitute the given values:

   $$N = \dfrac{N_0}{10^{\log R}}$$

5. Step 5 — Evaluate:

   $$N = 0.0054\ \text{org/L}$$

6. Step 6 — Check: returning N = 0.0054 org/L to

   $$\log_{10} R = \log_{10}\left(\dfrac{N_0}{N}\right)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0108 — kept a factor of two that cancels in the correct rearrangement.
• 0.0027 — dropped that same factor in the other direction.
• 0.0059 — rounded an intermediate value before the final step.

Reference: FE Handbook — Removal and Inactivation Requirements

removal and inactivation requirements for virus control under the surface water rule Given final pathogen concentration (N) = 77.1790 org/L; log removal/inactivation credit (logR) = 3.6200 log10, determine the initial pathogen concentration (N_0) in org/L.

Start with the thinking

• The governing relation printed in this handbook section is Removal and Inactivation Requirements.
• Everything except N_0 is given, so isolate N_0 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Removal and inactivation requirements set the required log reduction of pathogens between raw and finished drinking water.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\log_{10} R = \log_{10}\left(\dfrac{N_0}{N}\right)$$

2. Step 2 — Rearrange symbolically for N_0:

   $$N_{0} = N \times 10^{\log R}$$

3. Step 3 — List the givens: final pathogen concentration (N) = 77.1790 org/L, log removal/inactivation credit (logR) = 3.6200 log10.

4. Step 4 — Substitute the given values:

   $$N_{0} = 77.1790 \times 10^{\log R}$$

5. Step 5 — Evaluate:

   $$N_{0} = 321736\ \text{org/L}$$

6. Step 6 — Check: returning N_0 = 321,736 org/L to

   $$\log_{10} R = \log_{10}\left(\dfrac{N_0}{N}\right)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 643,471 — kept a factor of two that cancels in the correct rearrangement.
• 160,868 — dropped that same factor in the other direction.
• 353,909 — rounded an intermediate value before the final step.

Reference: FE Handbook — Removal and Inactivation Requirements

removal and inactivation requirements for Giardia in a filtration plant Given initial pathogen concentration (N_0) = 79,880 org/L; final pathogen concentration (N) = 47.3490 org/L, determine the log removal/inactivation credit (logR) in log10.

Start with the thinking

• The governing relation printed in this handbook section is Removal and Inactivation Requirements.
• Everything except logR is given, so isolate logR symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Removal and inactivation requirements set the required log reduction of pathogens between raw and finished drinking water.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\log_{10} R = \log_{10}\left(\dfrac{N_0}{N}\right)$$

2. Step 2 — Rearrange symbolically for logR:

   $$logR = \log_{10}\left(\dfrac{N_0}{N}\right)$$

3. Step 3 — List the givens: initial pathogen concentration (N_0) = 79,880 org/L, final pathogen concentration (N) = 47.3490 org/L.

4. Step 4 — Substitute the given values:

   $$logR = \log_{10}\left(\dfrac{N_0}{47.3490}\right)$$

5. Step 5 — Evaluate:

   $$logR = 3.2271\ \text{log10}$$

6. Step 6 — Check: returning logR = 3.2271 log10 to

   $$\log_{10} R = \log_{10}\left(\dfrac{N_0}{N}\right)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 6.4543 — kept a factor of two that cancels in the correct rearrangement.
• 1.6136 — dropped that same factor in the other direction.
• 3.5498 — rounded an intermediate value before the final step.

Reference: FE Handbook — Removal and Inactivation Requirements

removal and inactivation requirements log credit calculation Given initial pathogen concentration (N_0) = 47,100 org/L; log removal/inactivation credit (logR) = 1.9800 log10, determine the final pathogen concentration (N) in org/L.

Start with the thinking

• The governing relation printed in this handbook section is Removal and Inactivation Requirements.
• Everything except N is given, so isolate N symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Removal and inactivation requirements set the required log reduction of pathogens between raw and finished drinking water.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\log_{10} R = \log_{10}\left(\dfrac{N_0}{N}\right)$$

2. Step 2 — Rearrange symbolically for N:

   $$N = \dfrac{N_0}{10^{\log R}}$$

3. Step 3 — List the givens: initial pathogen concentration (N_0) = 47,100 org/L, log removal/inactivation credit (logR) = 1.9800 log10.

4. Step 4 — Substitute the given values:

   $$N = \dfrac{N_0}{10^{\log R}}$$

5. Step 5 — Evaluate:

   $$N = 493.2\ \text{org/L}$$

6. Step 6 — Check: returning N = 493.2 org/L to

   $$\log_{10} R = \log_{10}\left(\dfrac{N_0}{N}\right)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 986.4 — kept a factor of two that cancels in the correct rearrangement.
• 246.6 — dropped that same factor in the other direction.
• 542.5 — rounded an intermediate value before the final step.

Reference: FE Handbook — Removal and Inactivation Requirements

removal and inactivation requirements for virus control under the surface water rule Given final pathogen concentration (N) = 47.7710 org/L; log removal/inactivation credit (logR) = 4.7100 log10, determine the initial pathogen concentration (N_0) in org/L.

Start with the thinking

• The governing relation printed in this handbook section is Removal and Inactivation Requirements.
• Everything except N_0 is given, so isolate N_0 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Removal and inactivation requirements set the required log reduction of pathogens between raw and finished drinking water.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\log_{10} R = \log_{10}\left(\dfrac{N_0}{N}\right)$$

2. Step 2 — Rearrange symbolically for N_0:

   $$N_{0} = N \times 10^{\log R}$$

3. Step 3 — List the givens: final pathogen concentration (N) = 47.7710 org/L, log removal/inactivation credit (logR) = 4.7100 log10.

4. Step 4 — Substitute the given values:

   $$N_{0} = 47.7710 \times 10^{\log R}$$

5. Step 5 — Evaluate:

   $$N_{0} = 2449990\ \text{org/L}$$

6. Step 6 — Check: returning N_0 = 2,449,990 org/L to

   $$\log_{10} R = \log_{10}\left(\dfrac{N_0}{N}\right)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 4,899,980 — kept a factor of two that cancels in the correct rearrangement.
• 1,224,995 — dropped that same factor in the other direction.
• 2,694,989 — rounded an intermediate value before the final step.

Reference: FE Handbook — Removal and Inactivation Requirements

removal and inactivation requirements for Giardia in a filtration plant Given initial pathogen concentration (N_0) = 16,020 org/L; final pathogen concentration (N) = 39.3780 org/L, determine the log removal/inactivation credit (logR) in log10.

Start with the thinking

• The governing relation printed in this handbook section is Removal and Inactivation Requirements.
• Everything except logR is given, so isolate logR symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Removal and inactivation requirements set the required log reduction of pathogens between raw and finished drinking water.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\log_{10} R = \log_{10}\left(\dfrac{N_0}{N}\right)$$

2. Step 2 — Rearrange symbolically for logR:

   $$logR = \log_{10}\left(\dfrac{N_0}{N}\right)$$

3. Step 3 — List the givens: initial pathogen concentration (N_0) = 16,020 org/L, final pathogen concentration (N) = 39.3780 org/L.

4. Step 4 — Substitute the given values:

   $$logR = \log_{10}\left(\dfrac{N_0}{39.3780}\right)$$

5. Step 5 — Evaluate:

   $$logR = 2.6094\ \text{log10}$$

6. Step 6 — Check: returning logR = 2.6094 log10 to

   $$\log_{10} R = \log_{10}\left(\dfrac{N_0}{N}\right)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 5.2188 — kept a factor of two that cancels in the correct rearrangement.
• 1.3047 — dropped that same factor in the other direction.
• 2.8703 — rounded an intermediate value before the final step.

Reference: FE Handbook — Removal and Inactivation Requirements

removal and inactivation requirements log credit calculation Given initial pathogen concentration (N_0) = 75,360 org/L; log removal/inactivation credit (logR) = 2.9200 log10, determine the final pathogen concentration (N) in org/L.

Start with the thinking

• The governing relation printed in this handbook section is Removal and Inactivation Requirements.
• Everything except N is given, so isolate N symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Removal and inactivation requirements set the required log reduction of pathogens between raw and finished drinking water.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\log_{10} R = \log_{10}\left(\dfrac{N_0}{N}\right)$$

2. Step 2 — Rearrange symbolically for N:

   $$N = \dfrac{N_0}{10^{\log R}}$$

3. Step 3 — List the givens: initial pathogen concentration (N_0) = 75,360 org/L, log removal/inactivation credit (logR) = 2.9200 log10.

4. Step 4 — Substitute the given values:

   $$N = \dfrac{N_0}{10^{\log R}}$$

5. Step 5 — Evaluate:

   $$N = 90.6026\ \text{org/L}$$

6. Step 6 — Check: returning N = 90.6026 org/L to

   $$\log_{10} R = \log_{10}\left(\dfrac{N_0}{N}\right)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 181.2 — kept a factor of two that cancels in the correct rearrangement.
• 45.3013 — dropped that same factor in the other direction.
• 99.6629 — rounded an intermediate value before the final step.

Reference: FE Handbook — Removal and Inactivation Requirements

removal and inactivation requirements for virus control under the surface water rule Given final pathogen concentration (N) = 81.3390 org/L; log removal/inactivation credit (logR) = 3.8900 log10, determine the initial pathogen concentration (N_0) in org/L.

Start with the thinking

• The governing relation printed in this handbook section is Removal and Inactivation Requirements.
• Everything except N_0 is given, so isolate N_0 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Removal and inactivation requirements set the required log reduction of pathogens between raw and finished drinking water.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\log_{10} R = \log_{10}\left(\dfrac{N_0}{N}\right)$$

2. Step 2 — Rearrange symbolically for N_0:

   $$N_{0} = N \times 10^{\log R}$$

3. Step 3 — List the givens: final pathogen concentration (N) = 81.3390 org/L, log removal/inactivation credit (logR) = 3.8900 log10.

4. Step 4 — Substitute the given values:

   $$N_{0} = 81.3390 \times 10^{\log R}$$

5. Step 5 — Evaluate:

   $$N_{0} = 631392\ \text{org/L}$$

6. Step 6 — Check: returning N_0 = 631,392 org/L to

   $$\log_{10} R = \log_{10}\left(\dfrac{N_0}{N}\right)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 1,262,783 — kept a factor of two that cancels in the correct rearrangement.
• 315,696 — dropped that same factor in the other direction.
• 694,531 — rounded an intermediate value before the final step.

Reference: FE Handbook — Removal and Inactivation Requirements

removal and inactivation requirements for Giardia in a filtration plant Given initial pathogen concentration (N_0) = 64,620 org/L; final pathogen concentration (N) = 14.0750 org/L, determine the log removal/inactivation credit (logR) in log10.

Start with the thinking

• The governing relation printed in this handbook section is Removal and Inactivation Requirements.
• Everything except logR is given, so isolate logR symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Removal and inactivation requirements set the required log reduction of pathogens between raw and finished drinking water.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\log_{10} R = \log_{10}\left(\dfrac{N_0}{N}\right)$$

2. Step 2 — Rearrange symbolically for logR:

   $$logR = \log_{10}\left(\dfrac{N_0}{N}\right)$$

3. Step 3 — List the givens: initial pathogen concentration (N_0) = 64,620 org/L, final pathogen concentration (N) = 14.0750 org/L.

4. Step 4 — Substitute the given values:

   $$logR = \log_{10}\left(\dfrac{N_0}{14.0750}\right)$$

5. Step 5 — Evaluate:

   $$logR = 3.6619\ \text{log10}$$

6. Step 6 — Check: returning logR = 3.6619 log10 to

   $$\log_{10} R = \log_{10}\left(\dfrac{N_0}{N}\right)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 7.3238 — kept a factor of two that cancels in the correct rearrangement.
• 1.8310 — dropped that same factor in the other direction.
• 4.0281 — rounded an intermediate value before the final step.

Reference: FE Handbook — Removal and Inactivation Requirements