1. Lessons
  2. Gr12
  3. Math
  4. Advanced Functions
  5. Logarithms - Word Problems

Logarithms - Word Problems


The loudness, \(L\), of a rock concert is measured using the formula \(L = 10\log\left(\cfrac{I}{I_0}\right)\) where \(I\) is the intensity of sound power per unit area (\(\text{watts}/\text{m}^2\)) being measured and \(I_0\) is the intensity of sound power per unit area (\(\text{watts}/\text{m}^2\)) of the threshold of hearing.

How many times more loud is a rock concert with a sound intensity of \(123\;[\text{dB}]\) than the threshold of sound?

First, we can substitute \(L = 123\) into the equation:

\(L = 10\log\left(\cfrac{I}{I_0}\right)\)

\(123 = 10\log\left(\cfrac{I}{I_0}\right)\)

Next, we can divide both sides by \(10\):

\(12.3 = \log\left(\cfrac{I}{I_0}\right)\)

Then, we can rewrite the equation in exponential form. Once we do this, we are able to evaluate for the loudness of the rock concert relative to the threshold of sound:

\(\cfrac{I}{I_0} = 10^{12.3}\)

\(\cfrac{I}{I_0} = 1.99 \times 10^{12}\)

Therefore, we can determine that a rock concert with a sound intensity of \(123\;[\text{dB}]\) is \(\boldsymbol{1.99 \times 10^{12}}\) times louder than the threshold of hearing.

A radioactive substance has a half-life of \(3\) days. Suppose you have \(750 000\;[\text{g}]\) of this substance now. In how many days will the mass be \(2.3 \times 10^{-2}\;[\text{g}]\)?

First, we can identify what each value and variable represent:

  • \(750 000\) represents the initial mass of radioactive substance
  • \(0.5\) represents the decay factor
  • \(d\) represents the period in days where the radioactive substance mass decreases
  • \(3\) represents how often in days the radioactive substance halves in mass

Using all of the found values, we can write the equation as such:

\(y = 750 000(0.5)^{d/3}\)

Next, we can set the equation equal to \(2.3 \times 10^{-2}\) (or \(0.023\)). We can then divide both sides by \(750 000\):

\(0.023 = 750 000(0.5)^{d/3}\)

\(0.00000003 = (0.5)^{d/3}\)

After, we can rewrite the equation in logarithmic form:

\(\cfrac{d}{3} = \log_{0.5}(0.00000003)\)

Finally, we can use the Change of Base formula to convert the equation into a Natural Logarithm:

\(\cfrac{d}{3} = \cfrac{\ln(0.00000003)}{\ln(0.5)}\)

\(\cfrac{d}{3} = \cfrac{-17.30008955}{-0.69314718}\)

\(\cfrac{d}{3} = 24.95875342\)

\(d \approx 75\;[\text{days}]\)

Therefore, we can determine that it will take \(\boldsymbol{\approx 75\;[\textbf{days}]}\) for the mass to become \(2.3 \times 10^{-2}\;[\text{g}]\).

The pH scale is used to measure the acidity or alkalinity of a chemical solution. It is defined as \(pH= −\log[H^+]\) where \([H^+]\) is the concentration of hydronium ions, measured in moles per litre.

A fish tank's water was recently changed with distilled water of pH \(7\). The day after it was changed, apple juice was spilled into it which caused the pH to drop to \(5.8\). By what factor has \([H^+]\) changed?

First, we can write 2 equations, each representing the the respective pH levels of the different substances:

\(7= −\log(H^+_{\text{water}})\)

\(5.8= −\log(H^+_{\text{apple juice}})\)

Next, we can convert each equation to exponential form.

We can first convert the water equation:

\(-7= \log(H^+_{\text{water}})\)

\(10^{-7}= H^+_{\text{water}}\)

We can then convert the apple juice equation:

\(-5.8= \log(H^+_{\text{apple juice}})\)

\(10^{-5.8}= H^+_{\text{apple juice}}\)

Then, we can calculate the ratio of apple juice to regular water to determine the factor \([H^+]\) has changed:

\(= \cfrac{H^+_{\text{apple juice}}}{H^+_{\text{water}}}\)

\(= \cfrac{10^{-5.8}}{10^{-7}}\)

\(= 10^{-5.8 - (-7)}\)

\(= 10^{1.2}\)

\(= 15.8\)

Therefore, we can determine that \([H^+]\) has changed by a factor of \(\boldsymbol{15.8}\).

Find the decibel rating of a sound with an intensity of \(5000I_0\).

First, we can rewrite the equation and substitute \(I = 5000I_0\):

\(L = 10\log\left(\cfrac{I}{I_0}\right)\)

\(L = 10\log\left(\cfrac{5000I_0}{I_0}\right)\)

Next, we can evaluate the rest of the equation:

\(L = 10\log\left(\cfrac{5000\cancel{I_0}}{\cancel{I_0}}\right)\)

\(L = 10\log(5000)\)

\(L = 37\;[\text{dB}]\)

Therefore, we can determine the decibel rating of a sound with an intensity of \(5000I_0\) is \(\boldsymbol{37\;[\textbf{dB}]}\).

Review these lessons: