diff --git a/00016_Logarithm_Sound/en_article_proofreading.md b/00016_Logarithm_Sound/en_article_proofreading.md
index 771096b..0b58b67 100644
--- a/00016_Logarithm_Sound/en_article_proofreading.md
+++ b/00016_Logarithm_Sound/en_article_proofreading.md
@@ -6,36 +6,33 @@ keywords:
 is_finished: False
 ---
 
-# The sound
+# Sound
 
 Sound is a mechanical wave that we perceive by hearing.
-All people perceive the pitch and length of a tone in approximately the same way, 
+All people perceive the pitch and duration of a tone in approximately the same way, 
 but the perception of loudness is very subjective. 
-Loudness is determined by the amplitude of the oscillation of the environment, 
-in which the sound wave propagates. 
+Loudness is determined by the amplitude of the oscillation in the medium 
+through which the sound wave propagates. 
 Since the amplitude of sound waves is not easily measured, 
-the quantities sound intensity $I$ and the intensity sound level $L$
+the quantities of sound intensity $I$ and sound intensity level $L$
 are used to compare loudness objectively.
 
 *Sound intensity* expresses how much energy sound waves transfer to a unit area perpendicular to the direction of sound propagation per unit time.
-Healthy hearing can register the smallest sound intensity $I_0=10^{-12}\,\text{W}/\text{m}^2$ at a frequency of $1000\,\text{Hz}$, which corresponds to the hearing threshold.
-On the other hand, the sound intensity $10\,\text{W}/\text{m}^2$ is loud enough to correspond to the pain threshold.
+A healthy ear can detect the smallest sound intensity $I_0=10^{-12}\,\text{W}/\text{m}^2$ at a frequency of $1000\,\text{Hz}$, which corresponds to the hearing threshold.
+On the other hand, the sound intensity of $10\,\text{W}/\text{m}^2$ is loud enough to correspond to the pain threshold.
 However, increasing the sound intensity $I$ ten times does not correspond to perceiving the sound ten times louder. 
-Therefore, the sound intensity level $L$ is rather used to express sound volume,
+Therefore, the sound intensity level $L$ is rather used to express sound loudness,
 which uses a logarithmic scale in decibels (dB).
 
-*Sound intensity level* $L$ in decibels is defined by $$L=10\log{\frac{I}{I_0}},$$
+*Sound intensity level* $L$ in decibels is defined by the equation $$L=10\log{\frac{I}{I_0}},$$
 where $I$ is the sound intensity at the given location 
 and $I_0=10^{-12}\,\text{W}/\text{m}^2$, which corresponds to the hearing threshold.
-The sound intensity level $60\,\text{dB}$ corresponds to the volume of a normal conversation,
-$90\,\text{dB}$ is the volume of a lawnmower and $110\,\text{dB}$ 
-is the volume of a discotheque.
+The sound intensity level of $60\,\text{dB}$ corresponds to the loudness of a normal conversation, $90\,\text{dB}$ is the loudness of a lawnmower, and $110\,\text{dB}$ is the loudness of a discotheque.
+<!-- Termín “loudness” je přesnější a v tomto kontextu používanější než  “volume”, “Volume” má i další významy. Raději měním na “laudness”. -->
 There is a risk of hearing impairment for long-term listening 
 (even though we are not in pain) of volumes higher than $85\,\text{dB}$.
-From a volume higher than $100\,\text{dB}$ there is a risk of hearing damage within minutes.
-
-Let's explore the relation between sound intensity and sound intensity level, 
-i.e. the volume perceived by hearing.
+From a volume higher than $100\,\text{dB}$, there is a risk of hearing damage within minutes.
+ Let's explore the relation between sound intensity and sound intensity level, i.e., the loudness perceived by hearing.
 
 >**Exercise 1.** The sound intensity is $1{,}27\cdot10^{-3}\,\text{W}/\text{m}^2$ 
 >while listening to a speaker with a sound power of $20\,\text{W}$
@@ -43,17 +40,17 @@ i.e. the volume perceived by hearing.
 >(assume uniform transmission of the sound wave into free half-space).
 >How many decibels do we measure at this location?
 
-*Solution.* We will use the definition of sound intensity level $L=10\log{\frac{I}{I_0}},$ for calculation, where $I=1{,}27\cdot10^{-3}\,\text{W}/\text{m}^2$ is the sound intensity in
+*Solution.* We use the definition of sound intensity level $L=10\log{\frac{I}{I_0}},$ for calculation, where $I=1{,}27\cdot10^{-3}\,\text{W}/\text{m}^2$ is the sound intensity at the
 given location and $I_0=10^{-12}\,\text{W}/\text{m}^2$.
 $$L=10\log{\frac{I}{I_0}}=10\log{\left(\frac{1{,}27\cdot10^{-3}}{10^{-12}}\right)}=10\log{\left(1{,}27\cdot10^{9}\right)}=91\;\text{dB}\,.$$
 
-The sound intensity level is $91$ dB at a distancex $50$ m from the speaker, 
-which corresponds to the noise level of a motorcycle or lawnmower.
+The sound intensity level is $91$ dB at a distance $50$ m from the speaker, 
+which corresponds to the noise level of a motorcycle or a lawnmower.
 
->**Exercise 2.** How will the sound intensity level change if there will be
->twice the sound intensity at the location from the previous example, i.e. $2\cdot1{,}27\cdot10^{-3}\,\text{W}/\text{m}^2$ ?
+>**Exercise 2.** How will the sound intensity level change if there is
+>twice the sound intensity at the location from the previous example, i.e., $2\cdot1{,}27\cdot10^{-3}\,\text{W}/\text{m}^2$ ?
 
-*Solution.* We will use the same formula as in the last exercise:
+*Solution.* We use the same formula as in the previous exercise:
 $$L=10\log{\frac{I}{I_0}}=10\log{\left(\frac{2\cdot1{,}27\cdot10^{-3}}{10^{-12}}\right)}=10\log{\left(2{,}54\cdot10^{9}\right)}=94\;\text{dB}\,.$$
 The calculation shows that twice the sound intensity does not correspond 
 to twice the number of decibels. The sound intensity level will increase from
@@ -75,10 +72,10 @@ The sound intensity level increases by $3\,\text{dB}\,$ when the sound intensity
 >from the sound source is doubled? 
 
 *Solution.* Since the sound intensity $I$ is inversely proportional to the square of 
-distance, we get $$I=\frac{k}{l^2},$$ where $l$ is the distance from 
-the sound source. When the distance is doubled, the sound intensity
+the distance, we get $$I=\frac{k}{l^2},$$ where $l$ is the distance from 
+the sound source. When the distance is doubled, the sound intensity will be
 $$I=\frac{k}{(2l)^2}=\frac{1}{4}\cdot\frac{k}{l^2}\,.$$
-The sound intensity is reduced to $\frac{1}{4}$ of the original value.
+The sound intensity is reduced to $\frac{1}{4}$ of its original value.
 
 $$\Delta L=10\log{\frac{\frac{1}{4}I}{I_0}}-10\log{\frac{I}{I_0}}=10\cdot\left(\log{\frac{\frac{1}{4}I}{I_0}}-\log{\frac{I}{I_0}}\right)=10\log\left(\frac{\frac{\frac{1}{4}I}{I_0}}{\frac{I}{I_0}}\right)=10\log\frac{1}{4}=-6\;\text{dB}\,.$$
 The sound intensity level is reduced by $6\,\text{dB}$ if 
@@ -87,17 +84,18 @@ we double our distance from the sound source.
 >**Exercise 5.** From the formula for the sound intensity level $L=10\log{\frac{I}{I_0}}$,
 >express the sound intensity $I$.
 
-*Solution.* First, we have to isolate the logarithmic function 
-$\frac{L}{10}=\log{\frac{I}{I_0}}$ and then we will use the inverse 
-function for the logarithmic, i.e. exponential, function:
+*Solution.* First, we isolate the logarithmic function 
+$\frac{L}{10}=\log{\frac{I}{I_0}}$ and then we use the
+ inverse 
+function of the logarithm, i.e., the exponential function:
 $$10^{\frac{L}{10}}=\frac{I}{I_0}.$$
-From here we express the sound intensity
+From here, we express the sound intensity
 $$I=I_0 \cdot 10^{\frac{L}{10}}.$$
 
->**Exercise 6.** How many times will the sound intensity increase
+>**Exercise 6.** By how many times will the sound intensity increase
 >if the sound intensity level increases by $20\,\text{dB}$?
 
-*Solution.* The value of the sound intensity level changes from $L_1=L$ to $L_2=L+20\,\text{dB}$. We will use formula $I=I_0 \cdot 10^{\frac{L}{10}}$ 
+*Solution.* The value of the sound intensity level changes from $L_1=L$ to $L_2=L+20\,\text{dB}$. We use formula $I=I_0 \cdot 10^{\frac{L}{10}}$ 
 and express the quotient $\frac{I_2}{I_1}$:
 $$\frac{I_2}{I_1}=\frac{I_0 \cdot 10^{\frac{L_2}{10}}}{I_0 \cdot 10^{\frac{L_1}{10}}}=\frac{10^{\frac{L+20}{10}}}{10^{\frac{L}{10}}}=10^{\frac{L+20}{10}-\frac{L}{10}}=10^2=100.$$