In the same way that there are properties of exponents which allow us to simplify exponential expressions, there are properties of logarithms that allow us to simplify logarithmic expressions. In fact, each logarithm property is a consequence of an exponent property.

First, using the fact that \log_b (x)=n \iff x=b^n, it follows that: \begin{aligned}\log_b\left(b^x\right)=x&\text{ or }\ln e^x=x\\b^{\log_{b}x}=x &\text{ or }e^{\ln x}=x\end{aligned} Notice that the base of the exponent and the base of the logarithm are the same. This is called the inverse property of logarithms. We can also substitute x=0 and x=1 to get two special cases: \log_b\left(1\right)=0 \text{ or }\ln 1=0\\\log_b\left(b\right)=1\text{ or }\ln e=1

Since exponential and logarithmic equations are inverses of one another, we can derive logarithmic properties from the exponential properties.

	Exponent property	Logarithm property
Product property	b^{x}b^y=b^{x+y}	\log_b\left(xy\right) = \log_b\left(x\right)+\log_b\left(y\right)
Quotient property	\frac{b^x}{b^y}=b^{x-y}	\log_b\left(\frac{x}{y}\right) = \log_b\left(x\right)- \log_b\left(y\right)
Power property	\left(b^x\right)^y=b^{x\cdot y}	\log_b\left(x^p\right) = p\cdot \log_b\left(x\right)
Equality property	b^x=b^y \iff x=y	\log_b\left(x\right)=\log_b\left(y\right)\iff x=y

All properties associated with common logarithms can be applied to natural logarithms.

An additional property that helps simplify logarithmic expressions is the change of base property: \log_x\left(y\right) = \dfrac{\log_b\left(y\right)}{\log_b\left(x\right)}

The change of base property is best when we need to find a decimal approximation of a logarithm that cannot be simplified with the given properties. Most scientific calculators can only evaluate logarithms of base 10. We use this property to change the logarithm to base 10, then evaluate with a calculator.

Examples

Example 1

Rewrite the following logarithms without products, quotients, exponents, or radicals.

\log_3 \left(5\sqrt{x}\right)

Worked Solution

Create a strategy

The argument of the logarithm contains a product: 5\cdot \sqrt{x} This can be expanded using the product property of logarithms. We can also rewrite radicals with a rational exponent, then use the power property of logarithms to rewrite the expression without a power.

Apply the idea

\displaystyle \log_3 \left(5\sqrt{x}\right)	\displaystyle =	\displaystyle \log_3 5+\log_3 \sqrt{x}	Product property of logarithms
	\displaystyle =	\displaystyle \log_3 5+\log_3 x^{\frac{1}{2}}	Rewrite radical as rational exponent
	\displaystyle =	\displaystyle \log_3 5+\frac{1}{2}\log_3 x	Power property of logarithms

In the arguments, there are no more products or powers. Therefore, the expression has been fully expanded.

Reflect and check

Because the exponent only applied to the variable x, not to the coefficient of 5, the power only became the coefficient of \log_3 x, not the coefficient of \log_3 5. If the power applies to multiple terms, it will become the coefficient of the logarithms of each of those terms.

When expanding expressions with properties of logarithms, apply the product and quotient properties first, then the power property. This will help avoid making the mistake mentioned above.

\ln \left(\dfrac{4a}{9}\right)

Worked Solution

Create a strategy

The argument of the logarithm is a fraction which means expressions are being divided. We can expand these using the quotient property of logarithms.

There is also a product in the numerator of the argument which can be expanded with the product property of logarithms.

Apply the idea

\displaystyle \ln \left(\dfrac{4a}{9}\right)	\displaystyle =	\displaystyle \ln \left(4a\right)-\ln 9	Quotient property of logarithms
	\displaystyle =	\displaystyle \ln 4+\ln a-\ln 9	Product property of logarithms

Reflect and check

After the expression is fully expanded, any terms that were in the numerator of the original expression should be positive (product property) and any terms that were in the denominator of the original expression should be negative (quotient property).

Example 2

Rewrite the following expressions as a single logarithm.

3\log_3\left(x\right)-\log_3\left(4\right)-\log_3\left(x\right)

Worked Solution

Create a strategy

First, all terms in the expression have the same base. If any terms did not have the same base, we could not condense them into a single term.

When expanding expressions, we applied the product and quotient properties first, then the power property. When condensing into a single logarithm, we need to apply the properties backwards: power property first, then the product and quotient properties from left to right.

Apply the idea

\displaystyle 3\log_3\left(x\right)-\log_3\left(4\right)-\log_3\left(x\right)	\displaystyle =	\displaystyle \log_3\left(x^3\right)-\log_3\left(4\right)-\log_3\left(x\right)	Power property of logarithms
	\displaystyle =	\displaystyle \log_3\left(\frac{x^3}{4}\right)-\log_3\left(x\right)	Quotient property of logarithms
	\displaystyle =	\displaystyle \log_3\left(\frac{x^3}{4x}\right)	Quotient property of logarithms
	\displaystyle =	\displaystyle \log_3\left(\frac{x^2}{4}\right)	Simplify the argument

Reflect and check

The first term and the last term have the same base and the same argument. This makes them like terms, so we can combine them. 3\log_3\left(x\right)-\log_3\left(x\right)=2\log_3\left(x\right) This follows from the quotient property shown above. Both methods lead to the same answer and are valid methods of solving.

\displaystyle 3\log_3\left(x\right)-\log_3\left(4\right)-\log_3\left(x\right)	\displaystyle =	\displaystyle 2\log_3 \left(x\right)-\log_3\left(4\right)	Combine like terms
	\displaystyle =	\displaystyle \log_3 \left(x^2\right)-\log_3\left(4\right)	Power property of logarithms
	\displaystyle =	\displaystyle \log_3 \left(\frac{x^2}{4}\right)	Quotient property of logarithms

\ln x-5\ln y+\frac{1}{2}\ln z

Worked Solution

Create a strategy

When condensing terms into a single logarithm, we apply the power property before applying the product and quotient properties. The coefficient of a logarithm will become the exponent of its argument.

Apply the idea

\displaystyle \ln x-5\ln y+\frac{1}{2}\ln z	\displaystyle =	\displaystyle \ln x-\ln y^5+\ln z^{\frac{1}{2}}	Power property of logarithms
	\displaystyle =	\displaystyle \ln \left(\frac{x}{y^5}\right)+\ln z^{\frac{1}{2}}	Quotient property of logarithms
	\displaystyle =	\displaystyle \ln \left(\frac{xz^{\frac{1}{2}}}{y^5}\right)	Product property of logarithms
	\displaystyle =	\displaystyle \ln \left(\frac{x\sqrt{z}}{y^5}\right)	Rewrite rational exponent as radical

Reflect and check

When condensing logarithms, we always rewrite rational exponents as radicals. When expanding logarithms, we always rewrite radicals as a rational exponent and apply the power property.

Example 3

Evaluate the following expressions.

\log_{8}\left(16\right)-\log_{8}\left(2\right)

Worked Solution

Create a strategy

We can rewrite this expression using the quotient property of logarithms: \log_b\left(x\right)- \log_b\left(y\right) = \log_b\left(\frac{x}{y}\right)

Apply the idea

\displaystyle \log_{8}\left(16\right)-\log_{8}\left(2\right)	\displaystyle =	\displaystyle \log_{8}\left(\frac{16}{2}\right)	Quotient property of logarithms
	\displaystyle =	\displaystyle \log_8\left(8\right)	Evaluate the quotient
	\displaystyle =	\displaystyle 1	Inverse property of logarithms

Reflect and check

This example shows the importance of using logarithmic properties to simplify expressions. It would have been difficult to see how the original expression could be simplified, but the properties made the simplification much easier.

\log_{2}\left(3+\sqrt{5}\right)+\log_{2}\left(3-\sqrt{5}\right)

Worked Solution

Create a strategy

We can rewrite this expression using the product property of logarithms: \log_b\left(x\right)+ \log_b\left(y\right) =\log_b\left(xy\right)

Apply the idea

\displaystyle \log_{2}\left(3+\sqrt{5}\right)+\log_{2}\left(3-\sqrt{5}\right)	\displaystyle =	\displaystyle \log_{2}\left[\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)\right]	Product property of logarithms
	\displaystyle =	\displaystyle \log_2\left(9-5\right)	Distributive property
	\displaystyle =	\displaystyle \log_2\left(4\right)	Evaluate the difference
	\displaystyle =	\displaystyle \log_2\left(2^2\right)	Rewrite since 4=2^2
	\displaystyle =	\displaystyle 2\log_2\left(2\right)	Power property of logarithms
	\displaystyle =	\displaystyle 2\cdot 1	Inverse property of logarithms
	\displaystyle =	\displaystyle 2	Evaluate the product

Reflect and check

We could also solve this, once we arrive at \log_2\left(4\right), by observing that 4=2^2. We can then use the fact x=b^n \iff \log_b x=n to get \log_2\left(4\right)=2. In other words, the expression \log_2\left(4\right) is equal to the exponent that makes 2^x=4.

\displaystyle \log_{2}\left(3+\sqrt{5}\right)+\log_{2}\left(3-\sqrt{5}\right)	\displaystyle =	\displaystyle \log_{2}\left[\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)\right]	Product property of logarithms
	\displaystyle =	\displaystyle \log_2\left(9-5\right)	Distributive property
	\displaystyle =	\displaystyle \log_2\left(4\right)	Evaluate the difference
	\displaystyle =	\displaystyle 2	x=b^n \iff \log_b x=n

\log_{4}8

Worked Solution

Create a strategy

The argument of this logarithm is not a product or quotient and does not contain a power. In this case, notice that both 4 and 8 can be rewritten as base 2 to some power. We can use the change of base property to rewrite the expression in base 2, then use the inverse property to simplify.

Apply the idea

\displaystyle \log_4 8	\displaystyle =	\displaystyle \frac{\log_2 8}{\log_2 4}	Change of base property
	\displaystyle =	\displaystyle \frac{\log_2 2^3}{\log_2 2^2}	Rewrite since 4=2^2 and 8=2^3
	\displaystyle =	\displaystyle \frac{3}{2}	Inverse property of logarithms

Reflect and check

Alternatively, we could have changed these to base 10 and used a calculator to evaluate.

Idea summary

Properties of logarithms are derived from properties of exponents. They are as follows:

	Common log	Natural log
Inverse property	\log_b\left(b^x\right) = x, b^{\log_b x}=x	\ln \left(e^x\right)= x, e^{\ln x}=x
Product property	\log_b\left(xy\right) = \log_b\left(x\right)+\log_b\left(y\right)	\ln \left(xy\right)= \ln\left(x\right)+\ln\left(y\right)
Quotient property	\log_b\left(\frac{x}{y}\right) = \log_b\left(x\right)- \log_b\left(y\right)	\ln \left(\frac{x}{y}\right)= \ln\left(x\right)-\ln\left(y\right)
Power property	\log_b\left(x^p\right) = p\log_b\left(x\right)	\ln\left(x^p\right) = p\ln\left(x\right)
Equality property	\log_b\left(x\right)=\log_b\left(y\right)\iff \\x=y	\ln\left(x\right)=\ln\left(y\right)\iff \\x=y
Change of base property	\log_b\left(x\right) = \dfrac{\log_a\left(x\right)}{\log_a\left(b\right)}

Logarithm properties and transformations

The properties of logarithms have direct connections to transformations in the graph of a logarithmic function.

The product property, which states that log_b(xy) = log_b(x) + log_b(y), implies that a horizontal dilation of a logarithmic function, f(x) = log_b(kx), equates to a vertical translation. Essentially, this means multiplying the input by a constant (k) results in the graph shifting up or down by a distance related to the logarithm of that constant.

-4

-3

-2

-1

-4

-3

-2

-1

Consider the example: g\left(x\right)=\log_{2}\left(3x\right) shown on the graph along with the parent function f\left(x\right)=\log_{2}x. This representation clearly shows a horizontal dilation by a factor of 3.

Applying the product property to g\left(x\right):

g\left(x\right)=\log_{2}\left(3x\right)=\log_{2}3+ \log_{2}x

gives us a form of the function where we can clearly see a vertical translation of the parent function by \log_{2}3 units. However, this results in the same graph of g\left(x\right) showing these two transformations are equivalent.

The power property, which states that log_{b} x^n= n \cdot log_bx , tells us that raising the input of a logarithmic function to a power, f(x) = log_bx^k, results in a vertical dilation by a factor of k.

-4

-3

-2

-1

-4

-3

-2

-1

Consider the example: g\left(x\right)=\log_{2}x^3 shown on the graph along with the parent function f\left(x\right)=\log_{2}x.

Applying the power property to g\left(x\right):

g\left(x\right)=\log_{2}x^3=3 \cdot \log_{2}x

we can clearly see a vertical dilation of the parent function by a factor of 3, resulting in the same graph of g\left(x\right).

The change of base property, which states that log_{b} x= \dfrac{\log_{a}x}{log_{a}b}, demonstrates that all logarithmic functions are vertical dilations of each other.

-4

-3

-2

-1

-4

-3

-2

-1

Consider the example: g\left(x\right)=\log_{2}x shown on the graph.

Applying the change of base property to change the base from 2 to 3 we get:

g\left(x\right)=\log_{2}x=\dfrac{\log_{3}x}{\log_{3}2}=\dfrac{1}{\log_{3}2}\log_{3}x

we can clearly see that \log_{2}x is a vertical dilation of log_{3}x by a factor of \dfrac{1}{\log_{3}2}, resulting in the same graph of g\left(x\right).

Examples

Example 4

Given the logarithmic function f\left(x\right) = \log_{2}x, identify the vertical translation that is equivalent to the horizontal dilation f\left(x\right) = \log_{2}3x.

Worked Solution

Create a strategy

We can use the product property of logarithms, \log_{b}\left(xy\right) = \log_{b}\left(x\right) + \log_{b}\left(y\right), to transform the logarithmic function into a form that highlights the vertical translation.

Apply the idea

We start by applying the product rule to f\left(x\right) = \log_{2}x. This gives us f\left(x\right) = \log_{2}3 + \log_{2}x, or f\left(x\right) = a + \log_{2}x, where a = \log_{2}3.

Reflect and check

We see that the function has been vertically translated upwards by a = \log_{2}3. We can check this by plotting the original function and the transformed function on a graph and observing the vertical shift.

-4

-3

-2

-1

-4

-3

-2

-1

Idea summary

The properties of logarithms can be visually represented as transformations on the graph of a logarithmic function.

The product property corresponds to horizontal dilations and vertical translations.
The power property corresponds to vertical dilations.
The change of base property shows how all logarithmic functions can be considered as vertical dilations of each other.

2.12 Logarithmic function manipulation

Introduction

Ideas

Properties of logarithms

Examples

Example 1

Example 2

Example 3

Logarithm properties and transformations

Examples

Example 4

Outcomes

2.12.A

What is Mathspace

About Mathspace