Laws of logarithms

A logarithm is an exponent. So the laws of exponents give laws for logarithms. These laws turn products into sums and powers into products. This makes them the main tool for solving exponential equations.

The three laws

Let ${a>0}$ with ${a\ne 1}$, let ${A>0}$ and ${B>0}$, and let ${C}$ be any real number. Then:

Product law: ${\log_a(AB)=\log_a A+\log_a B}$

Quotient law: ${\log_a\dfrac{A}{B}=\log_a A-\log_a B}$

Power law: ${\log_a\left(A^C\right)=C\log_a A}$

In words: the log of a product is the sum of the logs. The log of a quotient is the difference of the logs. And an exponent inside a log can move to the front.

Why the product law is true. Let ${x=\log_a A}$ and ${y=\log_a B}$. Then ${A=a^x}$ and ${B=a^y}$. Multiply, and use a law of exponents:

${AB=a^x\cdot a^y=a^{x+y}}$

So ${\log_a(AB)=x+y=\log_a A+\log_a B}$. The other two laws come from ${\dfrac{a^x}{a^y}=a^{x-y}}$ and ${\left(a^x\right)^C=a^{Cx}}$ in the same way.

Example 1: Find each value.

(a) ${\log_4 2+\log_4 32}$

(b) ${\log_2 80-\log_2 5}$

Solution:

(a) Use the product law. Then ${4^3=64}$:

$\begin{align*}&\log_4 2+\log_4 32\\&=\log_4(2\cdot 32)\\&=\log_4 64=3\end{align*}$

(b) Use the quotient law. Then ${2^4=16}$:

$\begin{align*}&\log_2 80-\log_2 5\\&=\log_2\dfrac{80}{5}\\&=\log_2 16=4\end{align*}$

$\begin{align*}&-\dfrac{1}{3}\log 8\\&=\log 8^{-1/3}\\&=\log\dfrac{1}{2}\approx -0.301\end{align*}$

Expanding logarithms

To expand a logarithm means to write it as a sum and difference of simpler logarithms. Use the product and quotient laws first, and the power law last.

Example 2: Expand each logarithm. Assume all variables are positive.

(a) ${\log_2(6x)}$

(b) ${\log_5\left(x^3y^6\right)}$

Solution:

(a) ${\log_2(6x)=\log_2 6+\log_2 x}$

(b) Use the product law, then the power law:

$\begin{align*}&\log_5\left(x^3y^6\right)\\&=\log_5 x^3+\log_5 y^6\\&=3\log_5 x+6\log_5 y\end{align*}$

$\begin{align*}&\ln\dfrac{ab}{c^{1/3}}\\&=\ln(ab)-\ln c^{1/3}\\&=\ln a+\ln b-\dfrac{1}{3}\ln c\end{align*}$

Combining logarithms

To combine logarithms means the reverse: write a sum and difference of logarithms as a single logarithm. Use the power law first, to move each coefficient inside as an exponent. Then use the product and quotient laws.

Example 3: Write as a single logarithm.

(a) ${3\log x+\dfrac{1}{2}\log(x+1)}$

(b) $3\ln s+\dfrac{1}{2}\ln t-4\ln\left(t^2+1\right)$

Solution:

(a) Move the coefficients inside, then use the product law:

$\begin{align*}&3\log x+\dfrac{1}{2}\log(x+1)\\&=\log x^3+\log(x+1)^{1/2}\\&=\log\left(x^3\sqrt{x+1}\right)\end{align*}$

(b) Move the coefficients inside:

$\ln s^3+\ln t^{1/2}-\ln\left(t^2+1\right)^4$

The two added terms go in the numerator, and the subtracted term goes in the denominator:

$\ln\dfrac{s^3\sqrt{t}}{\left(t^2+1\right)^4}$

Watch out: There is no law for the log of a sum. ${\log(A+B)}$ is not ${\log A+\log B}$. For example, ${\log(1+1)=\log 2\approx 0.301}$, but ${\log 1+\log 1=0}$.

Also, ${\dfrac{\log A}{\log B}}$ is not ${\log\dfrac{A}{B}}$, and ${(\log A)^2}$ is not ${2\log A}$.

The change of base formula

Calculators have keys only for ${\log}$ (base ${10}$) and ${\ln}$ (base ${e}$). To find a logarithm with another base, change the base:

${\log_b x=\dfrac{\log_a x}{\log_a b}}$

Any base ${a}$ works on the right side. In practice, use ${\ln}$ or ${\log}$.

Why it is true. Let ${y=\log_b x}$. Then ${b^y=x}$. Take ${\log_a}$ of both sides, and use the power law:

$\begin{align*}\log_a b^y&=\log_a x\\y\log_a b&=\log_a x\end{align*}$

Divide by ${\log_a b}$. This gives the formula.

Example 4: Find each value to four decimal places.

(a) ${\log_8 5}$

(b) ${\log_2 10}$

Solution:

(a) Use natural logarithms:

$\log_8 5=\dfrac{\ln 5}{\ln 8}\approx\dfrac{1.6094}{2.0794}\approx 0.7740$

Check: ${8^{0.7740}\approx 5.00}$.

(b) Use common logarithms. Since ${\log 10=1}$:

$\log_2 10=\dfrac{\log 10}{\log 2}=\dfrac{1}{\log 2}\approx 3.3219$

This makes sense: ${10}$ is between ${2^3=8}$ and ${2^4=16}$.

An application

Example 5: The loudness of a sound, in decibels (dB), is

${L=10\log\dfrac{I}{I_0}}$.

Here ${I}$ is the intensity of the sound (the power it carries), and ${I_0}$ is the intensity of the faintest sound a person can hear. A rock concert is ${100{,}000}$ times as intense as normal talking. How many decibels louder is it?

Solution:

Let ${I_1}$ be the intensity of talking, and ${I_2=100{,}000\,I_1}$ the intensity of the concert. Subtract the two loudness values:

$\begin{align*}&L_2-L_1\\&=10\log\dfrac{I_2}{I_0}-10\log\dfrac{I_1}{I_0}\end{align*}$

Expand each log with the quotient law. The two ${\log I_0}$ terms cancel:

$\begin{align*}&L_2-L_1\\&=10\left(\log I_2-\log I_1\right)\\&=10\log\dfrac{I_2}{I_1}\end{align*}$

The ratio is ${\dfrac{I_2}{I_1}=100{,}000=10^5}$. So

${L_2-L_1=10\log 10^5=10\cdot 5=50}$.

The concert is ${50}$ dB louder than normal talking.

Summary

Product law: ${\log_a(AB)=\log_a A+\log_a B}$.
Quotient law: ${\log_a\dfrac{A}{B}=\log_a A-\log_a B}$.
Power law: ${\log_a A^C=C\log_a A}$.
There is no law for ${\log(A+B)}$.
Change of base: $\log_b x=\dfrac{\ln x}{\ln b}=\dfrac{\log x}{\log b}$.

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