Logarithmic Applications

Logarithms were invented to tame enormous ranges of numbers. When a quantity can vary by factors of millions or billions, a logarithmic scale compresses that range into manageable, human-readable values. This page covers the most important real-world logarithmic scales and shows you how to solve the equations that arise from them.

Why Logarithmic Scales Exist

Consider sound intensity. The faintest sound a human can hear has intensity about $10^{-12}$ watts per square meter. A jet engine at 30 meters has intensity about $10^2$ watts per square meter. That is a range of $10^{14}$ — a hundred trillion to one. Trying to compare these on a linear scale is impossible. But on a logarithmic scale, the faintest sound is 0 decibels and the jet is 140 decibels — a range from 0 to 140 that the human mind can grasp.

The general principle: when a quantity spans many orders of magnitude, use a log scale.

The Decibel Scale

The sound intensity level in decibels (dB) is defined as:

$L = 10 \log_{10}\left(\frac{I}{I_0}\right)$

where $I$ is the sound intensity (in W/m$^2$) and $I_0 = 10^{-12}$ W/m$^2$ is the reference intensity (threshold of hearing).

Key Properties

• Every 10 dB increase corresponds to a 10-fold increase in intensity
• Every 3 dB increase approximately doubles the intensity (since $10\log_{10}(2) \approx 3.01$)
• Decibels are additive for intensity ratios: if one source is 20 dB louder than another, it has $10^{20/10} = 100$ times the intensity

Example 1: A conversation has intensity $I = 10^{-6}$ W/m$^2$. What is the sound level in decibels?

$L = 10\log_{10}\left(\frac{10^{-6}}{10^{-12}}\right) = 10\log_{10}(10^6) = 10 \times 6 = 60 \text{ dB}$

Example 2: A rock concert is 115 dB. How many times more intense is it than normal conversation at 60 dB?

The difference is $115 - 60 = 55$ dB.

$\frac{I_{\text{concert}}}{I_{\text{conversation}}} = 10^{55/10} = 10^{5.5} \approx 316{,}228$

The concert is over 300,000 times more intense than conversation.

Example 3 (Reverse): An alarm has intensity $3.5 \times 10^{-4}$ W/m$^2$. Find its decibel level.

$L = 10\log_{10}\left(\frac{3.5 \times 10^{-4}}{10^{-12}}\right) = 10\log_{10}(3.5 \times 10^8)$

$= 10[\log_{10}(3.5) + \log_{10}(10^8)] = 10[0.544 + 8] = 10(8.544) = 85.44 \text{ dB}$

Application for Electricians

Electricians encounter decibels in signal processing and telecommunications. A cable that attenuates a signal by 6 dB reduces its power to:

$\frac{P_{\text{out}}}{P_{\text{in}}} = 10^{-6/10} = 10^{-0.6} \approx 0.251$

So only about 25 percent of the signal power passes through — a 6 dB loss means roughly one-quarter of the power remains.

The pH Scale

The pH of a solution measures its hydrogen ion concentration:

$\text{pH} = -\log_{10}[H^+]$

where $[H^+]$ is the hydrogen ion concentration in moles per liter (M).

Key Properties

• pH ranges from 0 (extremely acidic) to 14 (extremely basic) for most solutions
• pH 7 is neutral (pure water)
• Each 1-unit decrease in pH means a 10-fold increase in $[H^+]$ (more acidic)
• Each 1-unit increase in pH means a 10-fold decrease in $[H^+]$ (more basic)

Example 4: A solution has $[H^+] = 3.2 \times 10^{-4}$ M. Find its pH.

$\text{pH} = -\log_{10}(3.2 \times 10^{-4}) = -[\log_{10}(3.2) + \log_{10}(10^{-4})]$

$= -[0.505 + (-4)] = -(-3.495) = 3.495$

The pH is about 3.5 — acidic (similar to orange juice).

Example 5: A lake has pH 6.2 and a factory discharge has pH 3.8. How many times more concentrated is the hydrogen ion in the discharge?

$\frac{[H^+]_{\text{discharge}}}{[H^+]_{\text{lake}}} = \frac{10^{-3.8}}{10^{-6.2}} = 10^{6.2 - 3.8} = 10^{2.4} \approx 251$

The discharge is about 251 times more acidic.

Example 6 (Reverse): Find $[H^+]$ for blood with pH 7.4.

$[H^+] = 10^{-7.4} = 10^{-7} \times 10^{-0.4} \approx 10^{-7} \times 0.398 = 3.98 \times 10^{-8} \text{ M}$

Nursing Application

In nursing, blood pH is critical. Normal arterial blood pH is between 7.35 and 7.45. A pH below 7.35 is acidosis and above 7.45 is alkalosis. The pH difference between 7.35 and 7.45 corresponds to a hydrogen ion concentration ratio of $10^{0.1} \approx 1.26$ — a 26 percent change in $[H^+]$ is the entire normal range. This illustrates how sensitive biological systems are and why logarithmic measurement matters.

The Richter Scale

The Richter magnitude of an earthquake is:

$M = \log_{10}\left(\frac{A}{A_0}\right)$

where $A$ is the maximum amplitude of the seismograph trace and $A_0$ is a reference amplitude.

In terms of energy released, each whole-number increase in magnitude corresponds to approximately 31.6 times more energy (because the energy scale uses a factor of $10^{1.5} \approx 31.6$):

$\frac{E_2}{E_1} = 10^{1.5(M_2 - M_1)}$

Example 7: How much more energy does a magnitude 7.0 earthquake release compared to a magnitude 5.0?

$\frac{E_7}{E_5} = 10^{1.5(7 - 5)} = 10^{3} = 1000$

A magnitude 7 earthquake releases 1,000 times more energy than a magnitude 5.

Example 8: An earthquake releases 200 times more energy than a magnitude 4.0 event. What is its magnitude?

$200 = 10^{1.5(M - 4)}$

$\log_{10}(200) = 1.5(M - 4)$

$2.301 = 1.5(M - 4)$

$M - 4 = 1.534$

$M = 5.53$

Solving Real-World Logarithmic Equations

The general strategy for any applied log equation:

1. Isolate the logarithmic expression
2. Convert to exponential form: if $\log_b(x) = c$, then $x = b^c$
3. Solve the resulting equation
4. Check that solutions are in the domain (logarithms require positive arguments)

Example 9: An investment doubles when $2 = e^{0.05t}$. Solve for $t$.

$\ln 2 = 0.05t$

$t = \frac{\ln 2}{0.05} = \frac{0.693}{0.05} = 13.86 \text{ years}$

Example 10: A formula for loudness perception (in sones) is $S = 2^{(L - 40)/10}$, where $L$ is in decibels. If the perceived loudness doubles from 4 sones to 8 sones, by how many decibels did the sound increase?

$4 = 2^{(L_1 - 40)/10}$ and $8 = 2^{(L_2 - 40)/10}$

From $4 = 2^{(L_1-40)/10}$: since $4 = 2^2$, we get $(L_1 - 40)/10 = 2$, so $L_1 = 60$ dB.

From $8 = 2^{(L_2-40)/10}$: since $8 = 2^3$, we get $(L_2 - 40)/10 = 3$, so $L_2 = 70$ dB.

The increase is 10 dB — a doubling of perceived loudness corresponds to a 10 dB increase.

Interpreting Logarithmic Scales

When you see a graph with a logarithmic axis:

• Equal spacing represents equal ratios, not equal differences
• A straight line on a log-linear plot (log scale on $y$, linear on $x$) indicates exponential growth or decay
• A straight line on a log-log plot (both axes logarithmic) indicates a power-law relationship ($y = ax^n$)

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• Logarithmic scales compress huge ranges into manageable numbers — decibels for sound, pH for acidity, Richter for earthquakes
• The decibel scale: $L = 10\log_{10}(I/I_0)$, where each 10 dB = 10x intensity
• The pH scale: $\text{pH} = -\log_{10}[H^+]$, where each unit = 10x change in acidity
• The Richter scale: each whole-number increase = about 31.6x more energy
• To solve applied log equations: isolate, convert to exponential form, solve, and check domain
• A straight line on a log-linear plot means exponential; on a log-log plot means power-law

Return to College Algebra for more topics in this section.