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7. Inverse Functions and Logarithms
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United States of America
Grades 9-12

7.06 Properties of logarithms

Lesson
Practice

Introduction

In Math 2 lesson  1.02 Rational exponents  , we reviewed the integer exponent properties and extended them to rational exponents. Now, we will explore how these properties are related to properties of logarithms.

Properties of logarithms

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 propertyLogarithm property
Product propertyb^{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 propertyb^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.

a

\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 xPower 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.

b

\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 9Quotient property of logarithms
\displaystyle =\displaystyle \ln 4+\ln a-\ln 9Product 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.

a

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
b

\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.

a

\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 1Inverse 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.

b

\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 1Inverse property of logarithms
\displaystyle =\displaystyle 2Evaluate 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
c

\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 logNatural 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)}

Outcomes

F.LE.A.4

For exponential models, express as a logarithm the solution to ab^(ct) = d where a, c, and d are numbers and the base b is 2, 10, or e; evaluate the logarithm using technology.

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