Logarithmic Functions

Lesson

The addition property of logarithms relates the sum of two logarithms to the logarithm of a product.

Similarly, the subtraction property of logarithms relates the difference of two logarithms to the logarithm of a quotient.

Addition and subtraction property of logarithms

The addition property of logarithms is given by:

$\ln x+\ln y=\ln\left(xy\right)$`l``n``x`+`l``n``y`=`l``n`(`x``y`)

The subtraction property of logarithms is given by:

$\ln x-\ln y=\ln\left(\frac{x}{y}\right)$`l``n``x`−`l``n``y`=`l``n`(`x``y`)

We can prove these properties by using the corresponding properties of exponentials:

Proof of addition property:

We start by letting $\ln x=N$`l``n``x`=`N` and $\ln y=M$`l``n``y`=`M`. We can rewrite these two equations in their equivalent exponential forms, $e^N=x$`e``N`=`x` and $e^M=y$`e``M`=`y`. Multiplying these two expressions gives us the result:

$xy$xy |
$=$= | $e^N\times e^M$eN×eM |
(Writing down the product) |

$=$= | $e^{N+M}$eN+M |
(Using a property of exponentials) | |

$\ln\left(xy\right)$ln(xy) |
$=$= | $N+M$N+M |
(Rewriting in logarithmic form) |

$\ln\left(xy\right)$ln(xy) |
$=$= | $\ln x+\ln y$lnx+lny |
(Substituting) |

Proof of subtraction property:

We follow a similar procedure and start by letting $\ln x=N$`l``n``x`=`N` and $\ln y=M$`l``n``y`=`M`. We can rewrite these in their exponential forms, $e^N=x$`e``N`=`x` and $e^M=y$`e``M`=`y`. Taking the quotient of the two expressions gives us the result:

$\frac{x}{y}$xy |
$=$= | $\frac{e^N}{e^M}$eNeM |
(Writing down the quotient) |

$=$= | $e^{N-M}$eN−M |
(Using a property of exponentials) | |

$\ln\left(\frac{x}{y}\right)$ln(xy) |
$=$= | $N-M$N−M |
(Rewriting in logarithmic form) |

$\ln\left(\frac{x}{y}\right)$ln(xy) |
$=$= | $\ln x-\ln y$lnx−lny |
(Substituting) |

These two properties are especially valuable if we want to simplify expressions or solve equations involving logarithms.

Simplify the logarithmic expression $\ln12-\ln3$`l``n`12−`l``n`3.

**Think:** Since both logarithms are natural logarithms, they have the same base and so we can use the subtraction property of logarithms.

**Do:** To use the subtraction property of two logarithms, we can divide the arguments:

$\ln12-\ln3$ln12−ln3 |
$=$= | $\ln\left(\frac{12}{3}\right)$ln(123) |
(Using the subtraction property) |

$=$= | $\ln4$ln4 |
(Simplifying the argument) |

Rewrite $\ln5y$`l``n`5`y` as the sum or difference of two logarithms.

**Think:** Since there is a product in the logarithm, we can use the addition property in reverse.

**Do:** So using the addition property, we can rewrite $\ln5y$`l``n`5`y` in the form:

$\ln5+\ln y$`l``n`5+`l``n``y`

Rewrite $\ln3+\ln5$`l``n`3+`l``n`5 as a single logarithm.

Rewrite $\ln\left(\frac{19}{7}\right)$`l``n`(197) as a difference of logarithms.

The power property of logarithms allows us to rearrange logarithmic expressions in unique ways. We can use this to simplify expression, or more easily compare two equations.

Suppose we start with the quantity $\ln\left(x\right)$`l``n`(`x`) and we triple it. We can write the resulting quantity as $3\ln\left(x\right)$3`l``n`(`x`).

Next we can use repeated addition to write $3\ln\left(x\right)=\ln\left(x\right)+\ln\left(x\right)+\ln\left(x\right)$3`l``n`(`x`)=`l``n`(`x`)+`l``n`(`x`)+`l``n`(`x`). Now recall the following important property that is central to the idea of logarithms:

$\ln\left(a\right)+\ln\left(b\right)=\ln\left(ab\right)$`l``n`(`a`)+`l``n`(`b`)=`l``n`(`a``b`)

This says that the **sum** of the logarithm of two numbers is equal to the logarithm of the **product** of those two numbers. For the case of $3\ln\left(x\right)$3`l``n`(`x`), we have

$3\ln\left(x\right)$3ln(x) |
$=$= | $\ln\left(x\right)+\ln\left(x\right)+\ln\left(x\right)$ln(x)+ln(x)+ln(x) |
(Using repeated addition) |

$=$= | $\ln\left(xx\right)+\ln\left(x\right)$ln(xx)+ln(x) |
(Using the addition property) | |

$=$= | $\ln\left(xxx\right)$ln(xxx) |
(Using the addition property once more) | |

$=$= | $\ln\left(x^3\right)$ln(x3) |
(Writing repeated multiplication with a power) |

So we can see that $3\ln\left(x\right)=\ln\left(x^3\right)$3`l``n`(`x`)=`l``n`(`x`3). The same approach can be used for any other multiple of $\ln\left(x\right)$`l``n`(`x`), and we can summarise this result like so.

Power property of logarithms

For any number $a$`a`, we have

$\ln\left(x^a\right)=a\ln\left(x\right)$`l``n`(`x``a`)=`a``l``n`(`x`)

Once we are familiar with the power property, we can bring in our knowledge of other index properties and apply them to expressions involving logarithms.

Specifically, consider the following properties:

- $\frac{1}{x^a}=x^{-a}$1
`x``a`=`x`−`a` - $\sqrt[b]{x^a}=x^{\frac{a}{b}}$
^{b}√`x``a`=`x``a``b`

When we take the logarithm of these expressions, we get:

- $\ln\left(\frac{1}{x^a}\right)=\ln\left(x^{-a}\right)=-a\ln(x)$
`l``n`(1`x``a`)=`l``n`(`x`−`a`)=−`a``l``n`(`x`) - $\ln\left(\sqrt[b]{x^a}\right)=\ln\left(x^{\frac{a}{b}}\right)=\frac{a}{b}\ln(x)$
`l``n`(^{b}√`x``a`)=`l``n`(`x``a``b`)=`a``b``l``n`(`x`)

Let's look at an example that uses these properties.

Write the expression $\ln\left(\frac{1}{x^4}\right)$`l``n`(1`x`4) as a multiple of $\ln\left(x\right)$`l``n`(`x`).

**Think**: A multiple of $\ln\left(x\right)$`l``n`(`x`) means an expression of the form $a\ln\left(x\right)$`a``l``n`(`x`).

**Do**:

$\ln\left(\frac{1}{x^4}\right)$ln(1x4) |
$=$= | $\ln\left(x^{-4}\right)$ln(x−4) |

$=$= | $-4\ln\left(x\right)$−4ln(x) |

We have used the power property to write $\ln\left(\frac{1}{x^4}\right)$`l``n`(1`x`4) as $-4\ln\left(x\right)$−4`l``n`(`x`), which is a multiple of $\ln\left(x\right)$`l``n`(`x`).

Use the properties of logarithms to express $\ln y^{\frac{2}{9}}$`l``n``y`29 without any powers or surds.

Use the properties of logarithms to rewrite $-2\ln x$−2`l``n``x` with a positive power.

We've already seen the addition, subtraction and power properties of logarithms. We are able to use these properties in order to simplify or rewrite natural logarithm expressions. The natural logarithm is just $\log_ex$`l``o``g``e``x` represented as $\ln x$`l``n``x`.

Additional properties of the natural logarithm

$\ln\left(e\right)$ln(e) |
$=$= | $1$1 | The logarithm of the base is equal to one |

$\ln\left(1\right)$ln(1) |
$=$= | $0$0 | The logarithm of one is equal to zero |

The above properties can be used in order to simplify expressions, multiple properties may be used concurrently in order to simplify a single expression.

Careful!

Sometimes we are presented with a logarithm in a form where we can't immediately use logarithm properties, such as $\ln\sqrt{x}$`l``n`√`x`.

We can make use of exponent laws to first rewrite these types of expressions. For example, we can rewrite $\ln\sqrt{x}$`l``n`√`x` in the form $\ln\left(x^{\frac{1}{2}}\right)$`l``n`(`x`12). This now allows us to use properties of logarithms to further rewrite the expression.

Express $4\ln x+\ln y$4`l``n``x`+`l``n``y` as a single logarithm.

**Think:** Which logarithm properties can we identify within the above expression? Since there is a sum of multiple logarithms we want to use the addition property, but we can only do this if the two logarithms have the same coefficient. Using the power property on $4\ln x$4`l``n``x` first will result in both of the logarithms having the same coefficients.

**Do:** First let's remove the coefficient of $\ln x$`l``n``x` using the power property:

$4\ln x+\ln y$4lnx+lny |
$=$= | $\ln\left(x^4\right)+\ln y$ln(x4)+lny |

Now let's express the sum of the two logarithms as a single logarithm using the addition property:

$\ln\left(x^4\right)+\ln y$ln(x4)+lny |
$=$= | $\ln\left(x^4y\right)$ln(x4y) |

So, the simplest form of the sum of the two logarithms is $\ln\left(x^4y\right)$`l``n`(`x`4`y`).

Express $\ln\left(\frac{x}{y^3}\right)$`l``n`(`x``y`3) without using powers or fractions.

**Think:** Which logarithm properties can we identify within the above expression? Since there is a fraction within the logarithm we can use the subtraction property. The power of the variable $y$`y` means that we can also use the power property, but only once the single logarithm has been separated into two.

**Do:** First let's separate the logarithm into the difference of two logarithms using the subtraction property:

$\ln\left(\frac{x}{y^3}\right)$ln(xy3) |
$=$= | $\ln x-\ln\left(y^3\right)$lnx−ln(y3) |

Now let's remove the power in the logarithm $\ln\left(y^3\right)$`l``n`(`y`3) using the power property:

$\ln x-\ln\left(y^3\right)$lnx−ln(y3) |
$=$= | $\ln x-3\ln y$lnx−3lny |

So, the expression represented without any fractions or powers is $\ln x-3\ln y$`l``n``x`−3`l``n``y`.

Simplify $\frac{2\ln\left(x^3\right)}{6\ln\left(x^2\right)}$2`l``n`(`x`3)6`l``n`(`x`2).

**Think:** Which logarithm properties can we identify in the above expression? The exponents suggest that we can use the power property on each logarithm, and then simplify the remaining expression.

**Do:** First let's remove the power of $x$`x` in $2\ln\left(x^3\right)$2`l``n`(`x`3) using the power property:

$\frac{2\ln\left(x^3\right)}{6\ln\left(x^2\right)}$2ln(x3)6ln(x2) |
$=$= | $\frac{6\ln x}{6\ln\left(x^2\right)}$6lnx6ln(x2) |

Now let's remove the power in the logarithm in $6\ln\left(x^2\right)$6`l``n`(`x`2) using the power property:

$\frac{6\ln x}{6\ln\left(x^2\right)}$6lnx6ln(x2) |
$=$= | $\frac{6\ln x}{12\ln x}$6lnx12lnx |

We can now simplify by cancelling a common factor of $6$6 and $\ln x$`l``n``x` from the numerator and denominator:

$\frac{6\ln x}{12\ln x}$6lnx12lnx |
$=$= | $\frac{1}{2}$12 |

So the expression simplifies to $\frac{1}{2}$12.

Simplify $\frac{\ln a^6}{\ln a^3}$`l``n``a`6`l``n``a`3.

Express $6\ln x-\frac{1}{6}\ln y$6`l``n``x`−16`l``n``y` as a single logarithm.

Simplify the expression $\ln\left(x^2\right)+\ln\left(x^3\right)$`l``n`(`x`2)+`l``n`(`x`3), giving your answer in the form $k\ln x$`k``l``n``x`.

Manipulate rational, exponential, and logarithmic algebraic expressions

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